Question

Consider the following electrodes
P = Zn²⁺ (0.0001 M)/Zn  Q = Zn²⁺ (0.1 M)/Zn
R = Zn²⁺ (0.01 M)/Zn       S = Zn²⁺ (0.001 M)/Zn
E°Zn/Zn²⁺ = 0.76 V Electrode potentials of the above electrodes in volts are in the order

• A. P > S > R > Q
• B. S > R > Q > P
• C. Q > R > S > P
• D. P > Q > R > S

Answer 1

The Correct Option is C

We are given the following electrodes:
P = Zn²⁺ (0.0001 M) / Zn
Q = Zn²⁺ (0.1 M) / Zn
R = Zn²⁺ (0.01 M) / Zn
S = Zn²⁺ (0.001 M) / Zn

The standard electrode potential for Zn²⁺/Zn is \( E^0 = -0.76 \, \text{V} \).

Using the Nernst equation:
\[ E = E^0 + \frac{0.0592}{2} \log \left( \frac{[Zn^{2+}]_{\text{right}}}{[Zn^{2+}]_{\text{left}}} \right) \]

We calculate the electrode potentials for each electrode:

For P (Zn²⁺ = 0.0001 M):
\[ E_P = -0.76 + \frac{0.0592}{2} \log \left( \frac{1}{0.0001} \right) = -0.6416 \, \text{V} \]

For Q (Zn²⁺ = 0.1 M):
\[ E_Q = -0.76 + \frac{0.0592}{2} \log \left( \frac{1}{0.1} \right) = -0.7304 \, \text{V} \]

For R (Zn²⁺ = 0.01 M):
\[ E_R = -0.76 + \frac{0.0592}{2} \log \left( \frac{1}{0.01} \right) = -0.7008 \, \text{V} \]

For S (Zn²⁺ = 0.001 M):
\[ E_S = -0.76 + \frac{0.0592}{2} \log \left( \frac{1}{0.001} \right) = -0.6712 \, \text{V} \]

Therefore, the increasing order of electrode potentials is:
\[ S > R > Q > P \]

The correct answer is (C) : Q > R > S > P.

Answer 2

1. Recall the Nernst Equation
The Nernst equation relates the reduction potential of an electrochemical reaction to the standard electrode potential, temperature, and activities (or concentrations) of the chemical species involved:

$E = E^0 - \frac{RT}{nF} \ln Q$

where:
- $E$ is the cell potential
- $E^0$ is the standard cell potential
- $R$ is the ideal gas constant (8.314 J/(mol·K))
- $T$ is the temperature in Kelvin
- $n$ is the number of moles of electrons transferred in the cell reaction
- $F$ is Faraday's constant (96485 C/mol)
- $Q$ is the reaction quotient

We can also express the Nernst equation using the base-10 logarithm:

$E = E^0 - \frac{2.303RT}{nF} \log Q$

At 298 K (25°C), $\frac{2.303RT}{F} \approx 0.0592$ V, so the equation becomes:

$E = E^0 - \frac{0.0592}{n} \log Q$

2. Apply the Nernst Equation to the Zn²⁺/Zn Electrode
The half-cell reaction is: Zn²⁺ + 2e^- ⇌ Zn
So, $n = 2$.
The Nernst equation for this reaction is:

$E = E^0 - \frac{0.0592}{2} \log \frac{1}{[Zn^{2+}]}$
$E = E^0 + \frac{0.0592}{2} \log [Zn^{2+}]$

Here, $E^0 = -0.76$ V.

3. Calculate Electrode Potentials for P, Q, R, and S

Electrode P: $[Zn^{2+}] = 0.0001 M = 10^{-4} M$
$E_P = -0.76 + \frac{0.0592}{2} \log (10^{-4})$
$E_P = -0.76 + \frac{0.0592}{2} (-4)$
$E_P = -0.76 - 0.0592 \times 2 = -0.76 - 0.1184 = -0.8784 V$

Electrode Q: $[Zn^{2+}] = 0.1 M = 10^{-1} M$
$E_Q = -0.76 + \frac{0.0592}{2} \log (10^{-1})$
$E_Q = -0.76 + \frac{0.0592}{2} (-1)$
$E_Q = -0.76 - 0.0296 = -0.7896 V$

Electrode R: $[Zn^{2+}] = 0.01 M = 10^{-2} M$
$E_R = -0.76 + \frac{0.0592}{2} \log (10^{-2})$
$E_R = -0.76 + \frac{0.0592}{2} (-2)$
$E_R = -0.76 - 0.0592 = -0.8192 V$

Electrode S: $[Zn^{2+}] = 0.001 M = 10^{-3} M$
$E_S = -0.76 + \frac{0.0592}{2} \log (10^{-3})$
$E_S = -0.76 + \frac{0.0592}{2} (-3)$
$E_S = -0.76 - 0.0592 \times 1.5 = -0.76 - 0.0888 = -0.8488 V$

4. Arrange the Electrode Potentials in Descending Order
Comparing the values:
$E_Q = -0.7896 V$
$E_R = -0.8192 V$
$E_S = -0.8488 V$
$E_P = -0.8784 V$

Therefore, the order is: Q > R > S > P

5. Compare with the given options
Comparing the derived order with the options, we find that it matches option (C).

Final Answer:
(C) Q > R > S > P