Hi
Can someone please help me undersand how to get the answer to this question
A 15.0 mL solution of 0.050 mol/L AgNO3 was titrated with 0.0250 mol/L NaBr in the cell
SCE || titrated solution | Ag(s)
Find the cell voltage for 10.0 mL of titrant added.
(a) -0.0312 V
(b) 0.194 V
(c) 0.457 V
(d) 0.698 V
(e) None of these is correct
Check me out on this. The differing systems can be confusing.
The SCE placed on the left SHOULD mean it is the anode and the reaction occurring there is oxidation.
Look up your value of SCE, write it as an oxidation and the appropriate sign. I THINK it is -.241 if written as an oxidation.
For the right hand compartment of the cell, we have the reduction portion which is
E = Eo-(0.0592/n)*log[(Ag)/(Ag^+)]
E = 0.799 -(0.0592/1)log[(1)/(0.02)]
Check out reduction potential for Ag. It may be different from 0.799 in your text. The 0.02 comes from this.
15 mL x 0.05 = 0.75 mmoles Ag^+ initially.
10 mL x 0.025 = 0.25 mmoles Br^- added.
0.75-0.25 = 0.5 mmoles Ag^+ remaining in a volume of 25 mL which is 0.02 M.
Calculate E which is E of the Ag^+ cell.
Then Ecell = Eoxidization half + Eredn half = Ecalomel + EAg
Make sure the calomel is written as an oxidation and the sign is appropriate for the voltage.
To find the cell voltage for 10.0 mL of titrant added, we need to use the Nernst equation. The Nernst equation is given by:
Ecell = E°cell - (RT / nF) * ln(Q)
Where:
- Ecell is the cell voltage
- E°cell is the standard cell potential
- R is the gas constant (8.314 J/(mol·K))
- T is the temperature in Kelvin
- n is the number of electrons transferred in the balanced redox equation
- F is the Faraday constant (96,485 C/mol)
- Q is the reaction quotient
First, let's determine the balanced redox equation. The reaction between AgNO3 and NaBr can be represented as:
Ag+ + Br- --> AgBr(s)
This reaction involves the transfer of 1 electron, so n = 1.
Next, we need to calculate the reaction quotient, Q. The reaction quotient is determined using the concentrations of the species involved in the reaction. At the start of the titration, there is only AgNO3 in the cell. The concentration of AgNO3 is given as 0.050 mol/L.
Q = [Ag+] / [Br-]
Since Ag+ and Br- react in a 1:1 ratio, the concentration of Br- can be calculated using the volume and concentration of NaBr added. Given that 0.0250 mol/L NaBr was used and 10.0 mL of it was added, the moles of NaBr added can be calculated as follows:
moles of NaBr = concentration * volume
moles of NaBr = 0.0250 mol/L * (10.0 mL / 1000 mL/L)
moles of NaBr = 0.00025 mol
Since the reaction occurs in a 1:1 ratio, this also represents the moles of Br- ions produced.
Now we can calculate the concentration of Br-:
concentration of Br- = moles / volume
concentration of Br- = 0.00025 mol / (15.0 mL + 10.0 mL) / 1000 mL/L
concentration of Br- = 0.00025 mol / 25.0 mL / 1000 mL/L
concentration of Br- = 0.010 mol/L
Now that we have the concentrations of Ag+ and Br-, we can calculate the reaction quotient, Q:
Q = [Ag+] / [Br-]
Q = 0.050 mol/L / 0.010 mol/L
Q = 5
Now, we need to find the standard cell potential, E°cell. The standard cell potential can be found using the standard reduction potentials of the half-reactions involved in the cell. The half-reactions are:
Ag+ + e- --> Ag (reduction)
Br- --> 1/2 Br2 + e- (oxidation)
The reduction potential for Ag+ to Ag (E°red) is +0.800 V, and the reduction potential for Br- to Br2 (E°red) is +1.070 V.
Since the Ag half-reaction is the reduction, the reduction potential will be the positive value, and the Br half-reaction is the oxidation, so the reduction potential will be the negative value.
E°cell = E°red (reduction) - E°red (oxidation)
E°cell = +0.800 V - (+1.070 V)
E°cell = +0.800 V - 1.070 V
E°cell = -0.270 V
Finally, we can substitute these values into the Nernst equation to calculate the cell voltage (Ecell):
Ecell = E°cell - (RT / nF) * ln(Q)
Ecell = -0.270 V - ((8.314 J/(mol·K)) * (298 K) / (1 * 96,485 C/mol) * ln(5)
Calculating the above expression, we get approximately -0.0313 V.
Therefore, the cell voltage for 10.0 mL of titrant added is approximately -0.0312 V.
Thus, the correct option is (a) -0.0312 V.
To find the cell voltage for 10.0 mL of titrant added in this electrochemical cell, we need to understand the redox reaction that occurs during the titration.
The given cell is represented as:
SCE || titrated solution | Ag(s)
The SCE, or Standard Calomel Electrode, is the reference electrode. The line separating the two half-cells indicates a phase boundary, and Ag(s) represents the electrode where the reduction of silver ions occurs.
Let's break down the redox reaction that takes place in this cell:
AgNO3 (aq) + NaBr (aq) --> AgBr (s) + NaNO3 (aq)
In this reaction, silver ions (Ag+) from the silver nitrate (AgNO3) solution react with bromide ions (Br-) from the sodium bromide (NaBr) solution to form silver bromide (AgBr) precipitate. Sodium nitrate (NaNO3) is formed as a byproduct.
Now, to find the cell voltage, we need to consider the reduction potentials for the half-reactions taking place at each electrode.
The half-reaction at the Ag(s) electrode is the reduction of Ag+ ions:
Ag+ (aq) + e- --> Ag(s)
The reduction potential for this half-reaction is known as the standard reduction potential for silver ions (Ag+).
The half-reaction at the other electrode (the SCE) is not given explicitly. However, the SCE is a reference electrode with a known reduction potential of +0.241 V. Therefore, we can assume that the half-reaction at the SCE is the reduction of mercurous ions (Hg2+):
Hg2+ (aq) + 2e- --> 2Hg(l)
Now, to find the cell voltage, we need to calculate the difference between the reduction potentials of the two half-reactions:
Ecell = EAg+ - ESCE
Since EAg+ is not provided, we need to look up the standard reduction potential for silver ions (Ag+). It is available in reference tables, such as in a chemistry textbook or online.
Once you have the standard reduction potential for Ag+, subtract the reduction potential of the SCE (+0.241 V) from it to obtain the cell voltage.
Finally, calculate the number of moles of electrons involved in the reaction for the given volume of titrant added (10.0 mL), and multiply it by the cell voltage to get the cell voltage for the given volume.
Once you have obtained the cell voltage, compare it with the options provided (a, b, c, d, e) and select the correct one.
Note: It's essential to pay attention to the signs of the voltages and ensure that the reduction potentials are correctly adjusted for the balanced half-reactions.