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CBSE Class 12 Chemistry 2023 Delhi Set 1 Solved Paper

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Question : 19 of 35
Marks: +1, -0
SECTION - B
(a) (i) What should be the signs (positive/ negative) for ECellE^{\circ}_{\text{Cell}} and ΔG0\Delta G^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 K298\ \mathrm{K} :Fe(s)Fe2+(0.01 M)H(1M)+H2(g)(1 bar),\mathrm{Fe}_{(s)}|\mathrm{Fe}^{2+}(0.01\ \mathrm{M})||\mathrm{H}^{+}_{(1\mathrm{M})}|\mathrm{H}_{2(g)}(1\ \mathrm{bar}), Pt(s)\mathrm{Pt}_{(s)}
Given ECell=0.44 VE^{\circ}_{\text{Cell}} = 0.44\ \mathrm{V}.
(a) (i) ΔG0=veECell=+ve\begin{array}{l} \Delta G^0 = -\text{ve} \\ E^{\circ}_{\text{Cell}} = +\text{ve} \end{array} 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)+2H+(aq)Fe2+(aq)+H2(g)\mathrm{Fe}(\mathrm{s}) + 2 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow \mathrm{Fe}^{2+}(\mathrm{aq}) + \mathrm{H}_{2}(\mathrm{g})
By applying nearest Equation -
Ecell=Ecell0.0591nlogFe2+[H+]2E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.0591}{n} \log \frac{\mathrm{Fe}^{2+}}{[\mathrm{H}^{+}]^2}
=0.440.05912log0.0011=0.44 - \frac{0.0591}{2} \log \frac{0.001}{1}
=0.440.0295log103=0.44 - 0.0295 \log 10^{-3}
=0.440.0295(3log10)=0.44 - 0.0295(-3 \log 10) as, (log10=1)(\log 10 = 1)
=0.44+0.089=0.44 + 0.089
Ecell=0.53 VE_{\text{cell}} = 0.53\ \mathrm{V}
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