For the half cell reaction at 298 K- H - aq - e -1-2 H 2 g 1 atm A- E H - - H 20-59-2 - pH mVB- E H - - H 2-59-2 - pH mVC- E H - - H 20- pH -log- H - VD- E H - - H 2- pH -log- H - V

The correct option is B E_H^+/H_2 nbsp;= 59.2 ×pH mVH^+ (aq)+e^ →1/2 H_2 (g) E_H^+/H_2 = E_H^+/H_2^0 0.05921log(p_H_2) ^1/2[H^+] = 0+0.0592×log [H^+] ...

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Byju's Answer

Standard XII

Chemistry

Catalytic Hydrogenation

For the half ...

Question

For the half cell reaction at 298 K,
$H^{+} (a q) + e^{-} \to \frac{1}{2} H_{2} (g) (1 atm)$

E_{H^{+} / H_{2}} = - 59.2 \times p H m V

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E_{H^{+} / H_{2}}^{0} = - 59.2 \times p H m V

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E_{H^{+} / H_{2}}^{0} = p H \times log [H^{+}] V

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E_{H^{+} / H_{2}} = p H \times log [H^{+}] V

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Solution

The correct option is A $E_{H^{+} / H_{2}} = - 59.2 \times p H m V$
$H^{+} (a q) + e^{-} \to \frac{1}{2} H_{2} (g)$
$E_{H^{+} / H_{2}} = E_{H^{+} / H_{2}}^{0} - \frac{0.0592}{1} log \frac{(p_{H_{2}})^{\frac{1}{2}}}{[H^{+}]} = 0 + 0.0592 \times log [H^{+}] = - 0.0592 \times p H V = - 59.2 \times p H m V$

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For the half cell reaction at 298 K, H+(aq)+e−→12 H2(g) (1 atm)

The correct option is A EH+/H2=−59.2 ×pH mVH+(aq)+e−→12 H2(g) EH+/H2=E0H+/H2−0.05921log(pH2)12[H+]=0+0.0592×log [H+]=−0.0592×pH V=−59.2×pH mV

For the half cell reaction at 298 K,
$H^{+} (a q) + e^{-} \to \frac{1}{2} H_{2} (g) (1 atm)$