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Question: For the half cell reaction, \(\mathbf{H}^{\mathbf{+}}\mathbf{(aq) + e}\mathbf{\rightarrow}\frac{\mat...

For the half cell reaction, H+(aq)+e→12H2(g)1atm\mathbf{H}^{\mathbf{+}}\mathbf{(aq) + e}\mathbf{\rightarrow}\frac{\mathbf{1}}{\mathbf{2}}\underset{\mathbf{1atm}}{\mathbf{H}_{\mathbf{2}}\mathbf{(g)}}

A

E(H+/H2)=−59.2mV×pHE(H^{+}/H_{2}) = - 59.2mV \times pH

B

E0(H+/H2)=−59.2mV×pHE^{0}(H^{+}/H_{2}) = - 59.2mV \times pH

C

E0(H+/H2)=pHlog⁡[H+]E^{0}(H^{+}/H_{2}) = pH\log\lbrack H^{+}\rbrack

D

E(H+/H2)=−pHlog⁡[H+]E(H^{+}/H_{2}) = - pH\log\lbrack H^{+}\rbrack

Answer

E(H+/H2)=−59.2mV×pHE(H^{+}/H_{2}) = - 59.2mV \times pH

Explanation

Solution

EH+/H2=EH+/H20−0.05921log⁡PH21/2[H+]\mathbf{E}_{\mathbf{H}^{\mathbf{+}}\mathbf{/}\mathbf{H}_{\mathbf{2}}}\mathbf{=}\mathbf{E}_{\mathbf{H}^{\mathbf{+}}\mathbf{/}\mathbf{H}_{\mathbf{2}}}^{\mathbf{0}}\mathbf{-}\frac{\mathbf{0.0592}}{\mathbf{1}}\mathbf{\log}\frac{\mathbf{P}_{\mathbf{H}_{\mathbf{2}}}^{\mathbf{1/2}}}{\mathbf{\lbrack}\mathbf{H}^{\mathbf{+}}\mathbf{\rbrack}}

Because SHE =0; then, = 0+0.0592 log [H+]=−0.0592pH\lbrack H^{+}\rbrack = - 0.0592pH= – 59.2 mV × pH (mV = milli Volt)