Question

For the cell Cu(s)|Cu$^{2+}$(aq) (0.1 M)||Ag$^+$(aq) (0.01 M)|Ag(s) the cell potential E$_1$ = 0.3095 V
For the cell Cu(s)|Cu$^{2+}$(aq) (0.01 M)||Ag$^+$(aq) (0.001 M)|Ag(s) the cell potential = ________ $\times 10^{-2}$ V. (Round off to the Nearest Integer).
[Use : $\frac{2.303 RT}{F} = 0.059$]

Hint:

The Nernst equation is fundamental to electrochemistry. Remember to correctly identify the number of electrons transferred (n) and to raise the concentrations in the reaction quotient (Q) to the power of their stoichiometric coefficients.

Answer 1

Correct Answer: 28

Step 1: Determine the standard cell potential (E$^\circ_{cell}$) using the data for the first cell.
The overall cell reaction is: Cu(s) + 2Ag$^+$(aq) $\rightarrow$ Cu$^{2+}$(aq) + 2Ag(s).
The number of electrons transferred, n = 2.
The Nernst equation for the cell potential E is:
$E_{cell} = E^\circ_{cell} - \frac{0.059}{n} \log Q$, where $Q = \frac{[\text{Cu}^{2+}]}{[\text{Ag}^{+}]^2}$.
For the first cell (E$_1$):
$E_1 = 0.3095$ V, [Cu$^{2+}$] = 0.1 M, [Ag$^+$] = 0.01 M.
$0.3095 = E^\circ_{cell} - \frac{0.059}{2} \log\left(\frac{0.1}{(0.01)^2}\right)$.
$0.3095 = E^\circ_{cell} - 0.0295 \log\left(\frac{10^{-1}}{10^{-4}}\right) = E^\circ_{cell} - 0.0295 \log(10^3)$.
$0.3095 = E^\circ_{cell} - 0.0295 \times 3 = E^\circ_{cell} - 0.0885$.
$E^\circ_{cell} = 0.3095 + 0.0885 = 0.398$ V.
Step 2: Calculate the cell potential for the second cell (E$_2$) using the calculated E$^\circ_{cell}$.
For the second cell:
[Cu$^{2+}$] = 0.01 M, [Ag$^+$] = 0.001 M.
$E_2 = E^\circ_{cell} - \frac{0.059}{2} \log\left(\frac{[\text{Cu}^{2+}]}{[\text{Ag}^{+}]^2}\right)$.
$E_2 = 0.398 - 0.0295 \log\left(\frac{0.01}{(0.001)^2}\right)$.
$E_2 = 0.398 - 0.0295 \log\left(\frac{10^{-2}}{10^{-6}}\right) = 0.398 - 0.0295 \log(10^4)$.
$E_2 = 0.398 - 0.0295 \times 4 = 0.398 - 0.118$.
$E_2 = 0.280$ V.
Step 3: Express the answer in the required format.
The question asks for the potential in units of $\times 10^{-2}$ V.
$E_2 = 0.280 \text{ V} = 28.0 \times 10^{-2}$ V.
Rounding to the nearest integer, the value is 28.