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  • Calculate the reduction potential (in V) of the following electrode: <img alt="\mathrm{Pt}, \mathrm{H}_{2}(4 \mathrm{\: atm}) | \mathrm{\: H}_{2} \mathrm{SO}_{4}(0.01 \mathrm{\: M})" src="ht

Calculate the reduction potential (in V) of the following electrode:
\mathrm{Pt}, \mathrm{H}_{2}(4 \mathrm{\: atm}) | \mathrm{\: H}_{2} \mathrm{SO}_{4}(0.01 \mathrm{\: M}).
 

Option: 1

0.118


Option: 2

-0.118


Option: 3

0.059


Option: 4

-0.059


Answers (1)

best_answer

Half cell reaction:     


\mathrm{H}_{2}(4 \mathrm{\: atm}) \longrightarrow 2 \mathrm{H}^{+}(0.02 \mathrm{\: M})+2 \mathrm e^{-}
\begin{aligned} \mathrm E_{\left(\mathrm{H}_{2} | \mathrm{H}^{+}\right)} &=E_{\left(\mathrm{H}_{2} | \mathrm{H}^{+}\right)}^{-}-\frac{0.059}{2} \log \frac{\left[\mathrm{H}^{+}\right]^{2}}{\mathrm p_{\mathrm{H}_{2}}} \\ &=0-\frac{0.059}{2} \log \frac{(0.02)^{2}}{4 \mathrm{\: atm}} \\ &=-\frac{0.059}{2} \log \left(\frac{0.02 \times 0.02}{4}\right) \end{aligned}
\begin{aligned} &=-\frac{0.059}{2} \log \left(10^{-4}\right) \\ &=-\frac{0.059}{2} \times-4 \log 10 \\ &=0.059 \times 2=0.118 \mathrm{V} \\\\\ \mathrm E_{\left(\mathrm{H}^{+} | \mathrm{H}_{2}\right)} &=-0.118 \mathrm{\: V} \end{aligned}

 

Posted by

Shailly goel

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