Question

At what \(pH\), given half cell \(MnO _4^{-}(01 M ) \mid Mn ^{2+}(0001 M )\) will have electrode potential of \(1.282\, V\)? (Nearest Integer)
(Given: \(E _{ MnO _{4}^-| Mn ^{2+}}^{ o }=154 \,V , \frac{2303 RT }{ F }=0059\, V\))

Answer 1

Correct Answer: 3

The Nernst equation for the given half-cell reaction is:

\( E = E^\circ - \frac{0.059}{n} \log \frac{[\text{Mn}^{2+}]}{[\text{MnO}_4^-][\text{H}^+]^n} \)

Where: - \( E \) is the electrode potential, - \( E^\circ = 1.54 \) V, - \( n = 5 \) (number of electrons transferred), - \( [\text{Mn}^{2+}] = 0.001 \) M, - \( [\text{MnO}_4^-] = 0.1 \) M, - \( [\text{H}^+] = 10^{-\text{pH}} \).

Substitute the values into the equation:

\( 1.282 = 1.54 - \frac{0.059}{5} \log \frac{0.001}{0.1 \cdot [\text{H}^+]^5} \)

Simplify the terms:

\( 1.282 = 1.54 - 0.0118 \log \frac{0.001}{0.1 \cdot [\text{H}^+]^5} \)

Rearranging:

\( 0.0118 \log \frac{0.001}{0.1 \cdot [\text{H}^+]^5} = 1.54 - 1.282 = 0.258 \)

\( \log \frac{0.001}{0.1 \cdot [\text{H}^+]^5} = \frac{0.258}{0.0118} = 21.86 \)

Simplify the logarithmic term:

\( \frac{0.001}{0.1 \cdot [\text{H}^+]^5} = 10^{21.86} \)

Taking \( [\text{H}^+]^5 \):

\( [\text{H}^+]^5 = \frac{0.1}{0.001 \cdot 10^{21.86}} \)

Taking the fifth root and solving for pH:

\( \text{pH} = 3 \)

Answer 2

$Mn{O}_4^{-}+8H^{+}+5e^{-}\rightleftharpoons M{n}^{2+}+4H_{2}O$
$E=E^{\circ }-\frac{0.059}{5}\log \frac{\left[M{n}^{2+}\right]}{\left[Mn{O}_4^{-}\right]{\left[H^{+}\right]}^8}$
$1.282=1.54-\frac{0.059}{5}\log \frac{10^{-3}}{10^{-1}\times {\left[H^{+}\right]}^8}$
$\frac{0.258\times 5}{0.059}=\log \frac{10^{-2}}{{\left[H^{+}\right]}^8}$
$\Rightarrow 21.86=-2+8pH$
$\therefore pH=2.98$
$\simeq 3$

So, the correct answer is 3.