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(H2)=1 atm and [H+]=1.0 M (B) p(H2)=1 atm and [H+]=2.0 M (C) p(H2)=2 atm and [H+]=1.0 M (D) p(H2)=2 atm and [H+]=2.0 M

Tardigrade Admin · Accepted Answer

Reaction 2 H ++2 e arrow H 2 ∴ n=2, Q =(PH2/[H+]2) As EH- / H2 =EH- / H2° (-0.059/n) log Q =-(0.059/2) log Q ∴ if Q > 1 EH- / H2 < 0 A Q = (1/1) = 1 X B Q = (1/22) < 1 X C Q = (2/12) > 1 checkmark D Q = (2/22) < 1 X

A	$Q = \frac{1}{1} = 1$	X
B	$Q = \frac{1}{2 ^{2}} < 1$	X
C	$Q = \frac{2}{12} > 1$	$✓$
D	$Q = \frac{2}{2 ^{2}} < 1$	X

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

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

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

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

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

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

p(H2​)=1atm and [H+]=1.0M

p(H2​)=1atm and [H+]=2.0M

p(H2​)=2atm and [H+]=1.0M

p(H2​)=2atm and [H+]=2.0M

Reaction 2H++2e→H2​ ∴n=2,Q=[H+]2PH2​​​ As EH−/H2​​=EH−/H2​∘​n−0.059​logQ =−20.059​logQ ∴ if Q>1EH−/H2​​<0 A Q=11​=1 X B Q=221​<1 X C Q=122​>1 ✓ D Q=222​<1 X

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

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

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

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