Suppose that you want to construct a galvanic cell from nickel and cadmium. You use 0.01M sol of cadmium nitrate as one of the electrolytes. If you need a cell potential of 0.17V for this system, what concentration of NiCl2 should you use?
Please kindly specify your instructions or just provide the answer. thanks
Mama mo
Well, if you're feeling a bit charged up about constructing a galvanic cell, let's dive in, shall we?
To figure out the concentration of NiCl2, we can use the Nernst equation, which relates the cell potential (Ecell) to the concentrations of the reactants and products. In this case, we have:
Ecell = E°cell - (RT / nF) * ln(Q)
Where E°cell is the standard cell potential, R is the gas constant, T is the temperature in Kelvin, n is the number of moles of electrons transferred, F is Faraday's constant, and Q is the reaction quotient.
Since we want to calculate the concentration of NiCl2, we can rearrange the equation to solve for Q:
Q = e^((E°cell - Ecell) * nF / (RT))
Now, let's plug in the given values:
E°cell = 0.17V
Ecell = 0.17V
n = 2 (since 2 electrons are transferred in the reaction)
Let's assume the temperature is around 298 Kelvin. The gas constant (R) is approximately 8.314 J/(mol·K), and Faraday's constant (F) is about 96,485 C/mol.
Now, we need to figure out the value of Q using the equation:
Q = e^((E°cell - Ecell) * nF / (RT))
However, since we don't have the given values for the standard cell potential (E°cell) or the temperature (T), it's impossible to calculate the exact concentration of NiCl2 needed.
But hey, don't be too disappointed! Chemistry can be electrifying, even if we can't calculate everything with the information given. If you have any more questions, feel free to ask!
To calculate the concentration of NiCl2 you should use in order to achieve a specific cell potential of 0.17V, you need to consider the half-reactions involved in the cell and use the Nernst equation.
The half-reactions involved in this galvanic cell are:
1. Reduction half-reaction (Cathode): Cd2+(aq) + 2e- → Cd(s) (E° = -0.403V)
2. Oxidation half-reaction (Anode): Ni(s) → Ni2+(aq) + 2e- (E° = -0.257V)
Given that the desired cell potential (E°cell) is 0.17V, we can use the Nernst equation to relate the concentrations of the ions involved in the reaction with the cell potential:
Ecell = E°cell - (0.0592V/n)log(Q)
Where:
- Ecell is the cell potential
- E°cell is the standard cell potential (0.17V in this case)
- n is the number of moles of electrons transferred in the balanced equation (2 electrons in this case)
- Q is the reaction quotient (Cd2+/Ni2+ ratio)
To find the concentration of NiCl2 (Ni2+) needed, we need to solve the Nernst equation for Q, and then find the corresponding concentration of Ni2+:
0.17V = 0.17V - (0.0592V/2)log(Q)
Simplifying the equation, we get:
log(Q) = 0.0592V/2
Taking the antilog of both sides, we get:
Q = 10^(0.0592V/2)
Now, we can find the concentration of Cd2+ using the concentration of Cd(NO3)2 given (0.01M). In a 0.01M solution, the concentration of Cd2+ is also 0.01M.
Finally, we can use the ratio of Cd2+/Ni2+ (Q) to find the concentration of Ni2+:
Q = Cd2+/Ni2+
Cd2+ = 0.01M
Solving for Ni2+:
0.01M = Ni2+/Q
Ni2+ = 0.01M * Q
Substituting Q = 10^(0.0592V/2):
Ni2+ = 0.01M * 10^(0.0592V/2)
Now you can calculate the concentration of NiCl2 using the molar mass of NiCl2 and the molar mass of Ni2+ to convert from molarity to grams or moles.
Remember to round your final answer to an appropriate number of significant figures.
The cell reaction will be
Cd(s) + Ni^2+(aq) ==> Ni(s) + Cd(0.01M)
step 1. Calculate the E value for the Cd part using
E = Eo - (0.05916/n)*log (1/0.01) and the equation for the reduction of Cd^2+ to Cd(s) as in Cd^2+ + 2e ==> Cd(s). Now reverse the sign since you want the oxidation of Cd and not the reduction of Cd^2+.
Step 2. The half cells for the first reaction I gave above are
Cd(s)==> Cd^2+ + 2e .........E1 = from step 1
Ni^2+ + 2e ==> Ni(s)............E 2= ?
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Add to get Cd + Ni^2+ =>Cd^2+ + Ni E3 = 0.17 v.
Now calculate E2 = ? from E1 + E2 = E3
Step 3. Now you calculate the concentration of Ni^2+ from
Ni^2+ + 2e ==> Ni and E = Eo - (0.05916/n)*log[1/(Ni)].
Plug in Eo for Ni^2+ reduction, Eo is from the reduction tables, and E is the calculation from step 2 of E2. Solve for (Ni^2+).
Post ALL of your work if you get stuck.