CBSE Class 12 Chemistry 2023 Delhi Set 1 Solved Paper

Question : 19 of 35

Marks: +1, -0

SECTION - B

(a) (i) What should be the signs (positive/ negative) for

E°_{\text "Cell "}

and

\ΔG^0

for a spontaneous redox reaction occurring under standard conditions?
(ii) State Faraday's first law of electrolysis.
OR
(b) Calculate the emf of the following cell at

298 {K}

{Fe}_{({s})}|{Fe}^{2+}(0.01 {M})||{{H}^{+}}_{({1M})}| {H}_{2({g})}(1 {\bar}),

{Pt}_{({s})}

Given

{E}°_{{\Cell}} =0.44 {V}

Solution:

(a) (i)

\table \Δ{G}^0=-{ve};{E}°_{{\Cell}} =+ve|

for spon tan eous reaction
(ii) Faraday first law of electrolysis - The amount of chemical reaction which occur at any electrode during electrolysis by a current is proportional to the quantity of electricity passed through the electrolyte (solution).
OR
(b) According to the equation

{Fe}({s})+2 {H}^{+}({aq}) \→{Fe}^{+2}({aq})+{H}_2({g})

By applying nearest Equation -

{E}_{\text "cell "}={E}°_{\text "cell "} -{0.0591}/{{n}} \log {{Fe}^{2+}}/{[{H}^{+}]^2}

=0.44-{0.0591}/{2} \log {0.001}/{(1)}

=0.44-0.0295 \log 10^{-3}

=0.44-0.0295(-3 \log 10)

as,

(\log 10=1)

=0.44+0.089

{E}_{\text "cell "}=0.53 {V}

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