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.