Question

In which of the following cells, $E_{cell}$ = $E_{cell}^{\circ}$ ?

• A. Cu(s) | $Cu^{2+}$ (0.01M) || $Ag^{+}$ (0.1 M) | Ag(s)
• B. Pt($H_2$) | pH = 1 || $Zn^{2+}$ (0.1 M) | Zn(s)
• C. Pt($H_2$) | pH = 1 || $Zn^{2+}$ (1 M) | Zn (s)
• D. Pt($H_2$) | $H^{+}$ (0.01 M) || $Zn^{2+}$ (0.01 M) | Zn(s)

Answer 1

Answer: (A)

(1)

Answer 2

The relationship between the cell potential ($E_{cell}$) and the standard cell potential ($E_{cell}^{\circ}$) is given by the Nernst equation:

$E_{cell} = E_{cell}^{\circ} - \frac{RT}{nF} \ln Q$

For $E_{cell} = E_{cell}^{\circ}$, the term $\frac{RT}{nF} \ln Q$ must be zero. Since $R, T, n, F$ are non-zero constants (at a given temperature), this requires $\ln Q = 0$, which means $Q = 1$.

We need to find the cell among the given options where the reaction quotient $Q$ is equal to 1.

Let's determine the cell reaction and the expression for $Q$ for each cell, assuming the temperature is $298 K$ and the pressure of gases is 1 atm unless otherwise specified.

(1) Cu(s) | $Cu^{2+}$ (0.01M) || $Ag^{+}$ (0.1 M) | Ag(s)

Anode (oxidation): $Cu(s) \rightarrow Cu^{2+}(aq) + 2e^-$

Cathode (reduction): $Ag^{+}(aq) + e^- \rightarrow Ag(s)$

To balance the electrons transferred, we multiply the cathode reaction by 2:

$2 \times (Ag^{+}(aq) + e^- \rightarrow Ag(s)) \implies 2Ag^{+}(aq) + 2e^- \rightarrow 2Ag(s)$

Overall reaction: $Cu(s) + 2Ag^{+}(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s)$

The reaction quotient $Q$ is given by:

$Q = \frac{[Cu^{2+}]}{[Ag^{+}]^2}$

Given concentrations: $[Cu^{2+}] = 0.01$ M, $[Ag^{+}] = 0.1$ M.

$Q = \frac{0.01}{(0.1)^2} = \frac{0.01}{0.01} = 1$.

Since $Q=1$, for this cell, $E_{cell} = E_{cell}^{\circ}$.

(2) Pt($H_2$) | pH = 1 || $Zn^{2+}$ (0.1 M) | Zn(s)

The pH of the anode compartment is 1, so $[H^{+}] = 10^{-pH} = 10^{-1}$ M = 0.1 M.

Anode (oxidation): $H_2(g) \rightarrow 2H^{+}(aq) + 2e^-$

Cathode (reduction): $Zn^{2+}(aq) + 2e^- \rightarrow Zn(s)$

Overall reaction: $H_2(g) + Zn^{2+}(aq) \rightarrow 2H^{+}(aq) + Zn(s)$

Assuming standard pressure for $H_2$, $P_{H_2} = 1$ atm.

The reaction quotient $Q$ is given by:

$Q = \frac{[H^{+}]^2}{P_{H_2} [Zn^{2+}]}$

Given concentrations: $[H^{+}] = 0.1$ M, $[Zn^{2+}] = 0.1$ M. Assuming $P_{H_2} = 1$ atm.

$Q = \frac{(0.1)^2}{(1)(0.1)} = \frac{0.01}{0.1} = 0.1$.

Since $Q \neq 1$, $E_{cell} \neq E_{cell}^{\circ}$.

(3) Pt($H_2$) | pH = 1 || $Zn^{2+}$ (1 M) | Zn (s)

$[H^{+}] = 10^{-pH} = 10^{-1}$ M = 0.1 M.

Overall reaction: $H_2(g) + Zn^{2+}(aq) \rightarrow 2H^{+}(aq) + Zn(s)$

Assuming standard pressure for $H_2$, $P_{H_2} = 1$ atm.

$Q = \frac{[H^{+}]^2}{P_{H_2} [Zn^{2+}]}$

Given concentrations: $[H^{+}] = 0.1$ M, $[Zn^{2+}] = 1$ M. Assuming $P_{H_2} = 1$ atm.

$Q = \frac{(0.1)^2}{(1)(1)} = \frac{0.01}{1} = 0.01$.

Since $Q \neq 1$, $E_{cell} \neq E_{cell}^{\circ}$.

(4) Pt($H_2$) | $H^{+}$ (0.01 M) || $Zn^{2+}$ (0.01 M) | Zn(s)

Overall reaction: $H_2(g) + Zn^{2+}(aq) \rightarrow 2H^{+}(aq) + Zn(s)$

Assuming standard pressure for $H_2$, $P_{H_2} = 1$ atm.

$Q = \frac{[H^{+}]^2}{P_{H_2} [Zn^{2+}]}$

Given concentrations: $[H^{+}] = 0.01$ M, $[Zn^{2+}] = 0.01$ M. Assuming $P_{H_2} = 1$ atm.

$Q = \frac{(0.01)^2}{(1)(0.01)} = \frac{0.0001}{0.01} = 0.01$.

Since $Q \neq 1$, $E_{cell} \neq E_{cell}^{\circ}$.

Only in cell (1) is the reaction quotient $Q=1$. Therefore, in cell (1), $E_{cell} = E_{cell}^{\circ}$.