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

Question

The reduction potential of hydrogen half-cell will be negative if : (A) p ( H 2)=2 atm and [ H +]=2.0 M (B) p ( H 2)=1 atm and [ H +]=2.0 M (C) p ( H 2)=1 atm and [ H +]=1.0 M (D) p ( H 2)=2 atm and [ H +]=1.0 M

Tardigrade Admin · Accepted Answer

H ++ e - longrightarrow (1/2) H 2 Apply Nernst equation E =0 -(0.059/1) log ( P H 2(1/2)/[ H +]) E =-(0.059/1) log (21 / 2/1) Therefore E is negative.

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

$p (H_{2}) = 2 a t m$ and $[H^{+}] = 2.0 M$

$p (H_{2}) = 1 a t m$ and $[H^{+}] = 2.0 M$

$p (H_{2}) = 1 a t m$ and $[H^{+}] = 1.0 M$

$p (H_{2}) = 2 a t m$ and $[H^{+}] = 1.0 M$

$H^{+} + e^{-} ⟶ \frac{1}{2} H_{2}$

Apply Nernst equation

$E = 0 - \frac{0.059}{1} lo g \frac{P _{H_{2}}^{\frac{1}{2}}}{[ H ^{+} ]}$

$E = - \frac{0.059}{1} lo g \frac{2 ^{1/2}}{1}$

Therefore $E$ is negative.

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

p(H2​)=2atm and [H+]=2.0M

p(H2​)=1atm and [H+]=2.0M

p(H2​)=1atm and [H+]=1.0M

p(H2​)=2atm and [H+]=1.0M

H++e−⟶21​H2​ Apply Nernst equation E=0−10.059​log[H+]PH2​21​​​ E=−10.059​log121/2​ Therefore E is negative.

$p (H_{2}) = 2 a t m$ and $[H^{+}] = 2.0 M$

$p (H_{2}) = 1 a t m$ and $[H^{+}] = 2.0 M$

$p (H_{2}) = 1 a t m$ and $[H^{+}] = 1.0 M$

$p (H_{2}) = 2 a t m$ and $[H^{+}] = 1.0 M$

$H^{+} + e^{-} ⟶ \frac{1}{2} H_{2}$

Apply Nernst equation

$E = 0 - \frac{0.059}{1} lo g \frac{P _{H_{2}}^{\frac{1}{2}}}{[ H ^{+} ]}$

$E = - \frac{0.059}{1} lo g \frac{2 ^{1/2}}{1}$

Therefore $E$ is negative.