Question

The electrode potential of electrode

$\text{M }(s) \rightarrow M^{\text{n+}}\ (\text{aq})\ (\text{2M})\text{ + n}\text{e}^{-}.$ at 298 K is E1. When temperature is doubled and concentration is made half, then the electrode potential becomes E2. Which of the following represents the correct relationship between E1 and E2 ?

• A. $E_{1}\text{ > }\text{E}_{2}$
• B. $E_{1}\text{ < }\text{E}_{2}$
• C. $E_{1}\text{ = }\text{E}_{2}$
• D. Cann’t be predicted

Answer 1

Answer: (B)

$E_{1}\text{ < }\text{E}_{2}$

Answer 2

$E_{1}\text{ = }\text{E}^{0}\text{ -}\frac{\text{RT}}{\text{nF}}\text{ln 2 }$

$E_{2}\text{ = }\text{E}^{0}\text{ - }\frac{R \times \text{2T}}{\text{nF}}\text{ln1 = }\text{E}^{0}$

$\therefore$ E2 > E1