Consider the above electrochemical cell where a metal electrode (M) is undergoing redox reaction by forming $M^{+} (M \to M^{+} + e^{-})$ . The cation $M^{+}$ is present in two different concentrations $c_{1} and c_{2}$ as shown above. Which of the following statement is correct for generating a positive cell potential [2026]

Question

Smart Achievers · Accepted Answer

(4)

$M^{+} (M \to M^{+} + e^{-})$

$M^{+} (C_{2}) \to M^{+} (C_{1})$

$E_{cell} = - 0.059 \log \frac{C_{1}}{C_{2}}$

$∴$ $E_{cell} > 0 \Rightarrow C_{1} < C_{2}$

$(2) If C_{1} is at cathode$

$M^{+} (C_{1}) \to M^{+} (C_{2})$

$E_{cell} = - 0.059 \log \frac{C_{2}}{C_{1}} > 0$

$C_{2} < C_{1}$

Solution

1 If $c_{1}$ is present at cathode, then $c_{1} < c_{2}$

2 If $c_{1}$ is present at anode, then $c_{1} > c_{2}$

3 If $c_{1}$ is present at anode, then $c_{1} = c_{2}$

4 If $c_{1}$ is present at cathode, then $c_{1} > c_{2}$ .

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Solution

1 If c1 is present at cathode, then c1<c2

2 If c1 is present at anode, then c1>c2

3 If c1 is present at anode, then c1=c2

4 If c1 is present at cathode, then c1>c2.

Ans. (4) (1) If C1 is at anode ⇒ cell reaction M+(C2)→M+(C1) Ecell=-0.059logC1C2 ∴ Ecell>0⇒C1<C2 (2) If C1 is at cathode M+(C1)→M+(C2) Ecell=-0.059logC2C1>0 C2<C1

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1 If $c_{1}$ is present at cathode, then $c_{1} < c_{2}$

2 If $c_{1}$ is present at anode, then $c_{1} > c_{2}$

3 If $c_{1}$ is present at anode, then $c_{1} = c_{2}$

4 If $c_{1}$ is present at cathode, then $c_{1} > c_{2}$ .