(C) The half-cell reaction is $2H^{+} + 2e^{-} \rightleftharpoons H_2$. Using the Nernst equation: $E = E^{\circ} - \frac{0.059}{n} \log \frac{P_{H_2}

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(C) The half-cell reaction is $2H^{+} + 2e^{-} \rightleftharpoons H_2$. Using the Nernst equation: $E = E^{\circ} - \frac{0.059}{n} \log \frac{P_{H_2}}{[H^{+}]^2}$. Given $E = -0.059 \ V$,$E^{\circ} = 0 \ V$ (for $SHE$),$n = 2$,and $P_{H_2} = 1 \ bar$. Substituting the values: $-0.059 = 0 - \frac{0.059}{2} \log \frac{1}{[H^{+}]^2}$. $-0.059 = -\frac{0.059}{2} \times (-2 \log [H^{+}])$. $-0.059 = 0.059 \log [H^{+}]$. $\log [H^{+}] = -1$. $[H^{+}] = 10^{-1} = 0.1 \ M$.

Class 12 Chemistry Electrochemistry Nernst equation and ECS

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Consider the single electrode process $2H^{+} + 2e^{-} \rightleftharpoons H_2$ catalyzed by a platinum black electrode in $HCl$ electrolyte. The potential of the electrode is $-0.059 \ V$. What is the concentration of the acid in the hydrogen half-cell if the $H_2$ pressure is $1 \ bar$ (in $M$)?

TS EAMCET 2017 All TS EAMCET

A
$1$
B
$10$
C
$0.1$
D
$0.01$

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Consider the single electrode process $2H^{+} + 2e^{-} \rightleftharpoons H_2$ catalyzed by a platinum black electrode in $HCl$ electrolyte. The potential of the electrode is $-0.059 \ V$. What is the concentration of the acid in the hydrogen half-cell if the $H_2$ pressure is $1 \ bar$ (in $M$)?

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Which of the following statements is true regarding the cell emf at $298 \ K$ for the cell $Ni_{(s)} | Ni^{2+}(0.01 \ M) || Ag^{+}(0.01 \ M) | Ag_{(s)}$?

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What will be the oxidation potential for the following hydrogen half cell at $1 \ bar$ pressure and $25^{\circ} C$ temperature (in $V$)?
$Pt \mid H_{2(g)} (1 \ bar) \mid HCl_{(aq)} \ pH = 3$