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Question

Chemistry Question on Thermodynamics

The reduction potential of hydrogen half-cell will be negative if :

A

p(H2)=2atmp \left( H _{2}\right)=2 atm and [H+]=2.0M\left[ H ^{+}\right]=2.0 M

B

p(H2)=1atmp \left( H _{2}\right)=1 atm and [H+]=2.0M\left[ H ^{+}\right]=2.0 M

C

p(H2)=1atmp \left( H _{2}\right)=1 atm and [H+]=1.0M\left[ H ^{+}\right]=1.0 M

D

p(H2)=2atmp \left( H _{2}\right)=2 atm and [H+]=1.0M\left[ H ^{+}\right]=1.0 M

Answer

p(H2)=2atmp \left( H _{2}\right)=2 atm and [H+]=1.0M\left[ H ^{+}\right]=1.0 M

Explanation

Solution

H++e12H2H ^{+}+ e ^{-} \longrightarrow \frac{1}{2} H _{2} Apply Nernst equation E=00.0591logPH212[H+]E =0 -\frac{0.059}{1} \log \frac{ P _{ H _{2}}^{\frac{1}{2}}}{\left[ H ^{+}\right]} E=0.0591log21/21E =-\frac{0.059}{1} \log \frac{2^{1 / 2}}{1} Therefore EE is negative.