A galvanic cell is constructed with a silver-silver chloride electrode, and a nickel strip immersed in a beaker containing 6.96 x 10-2 M solution of NiCl2. Determine the balanced cell reaction and calculate the potential of the cell. Enter in volts. (assume a temperature of 25°C)
First note that the Ni electrode is not in 1 M solution; therefore, the E value is not Eo. That must be corrected.
E = Eored - 0.RT/nF*[log (red form)/(oxidized form)]
Plug in the values. -0.230 - (0.0592/2)log (Ni)/(Ni^2+). Plug in 1 for Ni and 0.0696 for (Ni^2+) and solve for Ered. Reverse that for Eox.
Ag^+ + e ==> Ag(s) Eored = +0.230
Ni ==> Ni^2+ + 2e Eoox = value from above.
---------------------------------
2Ag + Ni(s) => 2Ag(s) + Ni^2+ and Ecell = Ered + Eox
Note also: The values I found for E for Ag/AgCl on the web were not consistent. Use the values in your text or notes for this problem and not the numbers I've used.
To determine the balanced cell reaction and calculate the potential of the cell, we need to follow these steps:
Step 1: Write the half-reactions
The anode half-reaction occurs at the nickel strip, which will oxidize:
Ni(s) → Ni2+(aq) + 2e-
The cathode half-reaction occurs at the silver-silver chloride electrode, where silver ion is reduced:
AgCl(s) + e- → Ag(s) + Cl-(aq)
Step 2: Balance the half-reactions
To balance the half-reactions, we must ensure that the number of electrons produced in the oxidation half-reaction equals the number of electrons consumed in the reduction half-reaction.
For the nickel-half reaction, there are already 2 electrons on the left side. Therefore, no further balancing is needed.
For the silver-silver chloride electrode half-reaction, we need to balance the charges:
AgCl(s) + e- → Ag(s) + Cl-(aq)
Step 3: Combine the half-reactions
The overall balanced cell reaction can be obtained by adding the two balanced half-reactions together:
Ni(s) + 2AgCl(s) → Ni2+(aq) + 2Ag(s) + 2Cl-(aq)
Step 4: Calculate the cell potential
The standard cell potential can be calculated using the Nernst equation:
Ecell = E°cell - (0.0592 V / n) * log(Q)
In this case, we are given the concentration of NiCl2 as 6.96 x 10^-2 M. Since there are no concentrations given for AgCl, we assume its concentration to be 1 M.
The reaction quotient (Q) can be calculated as:
Q = [Ni2+] / [Ag+]^2 * [Cl-]^2
Using the given concentrations and assuming 1 M for AgCl, we can substitute the values into the equation.
Step 5: Calculate the cell potential
The standard cell potential (E°cell) for the reaction can be looked up in a table as 0.799 V.
Using the Nernst equation, we can calculate the cell potential (Ecell) at 25°C.
Let's calculate Ecell using the equation: Ecell = E°cell - (0.0592 V / n) * log(Q)
Assume n = 2 for the balanced half-reaction, we have:
Ecell = 0.799 V - (0.0592 V / 2) * log(Q)
Substitute the value of Q into the equation and calculate Ecell.
To determine the cell potential, we need to first determine the balanced cell reaction. Here's how you can do it step by step:
Step 1: Write the half-reactions
The silver-silver chloride electrode is the cathode (reduction half-reaction), and the nickel strip is the anode (oxidation half-reaction).
Cathode (reduction half-reaction):
AgCl(s) + e- -> Ag(s) + Cl-
Anode (oxidation half-reaction):
Ni(s) -> Ni2+(aq) + 2e-
Step 2: Balance the half-reactions
To balance the half-reactions, make sure that the number of electrons in each half-reaction is equal. Multiply the reduction and oxidation half-reactions by appropriate coefficients to achieve this.
Cathode (reduction half-reaction):
2AgCl(s) + 2e- -> 2Ag(s) + 2Cl-
Anode (oxidation half-reaction):
Ni(s) -> Ni2+(aq) + 2e-
Step 3: Combine the half-reactions
Add the balanced reduction and oxidation half-reactions to obtain the overall balanced cell reaction.
Cell reaction:
2AgCl(s) + Ni(s) -> 2Ag(s) + Ni2+(aq) + 2Cl-
Step 4: Calculate cell potential
Now that we have the balanced cell reaction, we can use the Standard Reduction Potentials table to find the standard reduction potentials (E°) for each half-reaction.
AgCl(s) + e- -> Ag(s) + Cl- : E° = +0.222 V
Ni2+(aq) + 2e- -> Ni(s) : E° = -0.257 V
To calculate the cell potential (Ecell), use the equation:
Ecell = E°cathode - E°anode
Ecell = (+0.222 V) - (-0.257 V)
Ecell = +0.479 V
Therefore, the potential of the cell is +0.479 volts at 25°C.