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Chemistry · 2026 · 5 marks
CBSE 2026 · Region 4 · Set 1 · Q33
(a) (i) Calculate E_cell^0 for the following reaction which is at equilibrium: Cu ( s )+2 Ag^+( aq ) ⇌ Cu^2+( aq )+2 Ag ( s ) Equilibrium constant (…
(i)
Calculate $\displaystyle \mathrm{E}_{\text {cell }}^{0}$ for the following reaction which is at equilibrium : $\displaystyle \mathrm{Cu}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) \rightleftharpoons \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})$ Equilibrium constant $\displaystyle \left(\mathrm{K}_{\mathrm{c}}\right)$ for the cell is $\displaystyle 10^{15}$. [Given : $\displaystyle \log 10=1$ ]
(ii)
Write anode, cathode and overall reaction of lead storage battery when it is in use.
(iii)
How much electricity is required in coulombs for the oxidation of $\displaystyle 1$ mol of FeO to $\displaystyle \mathrm{Fe}_{2} \mathrm{O}_{3}$ ?
(i)
Conductivity of $\displaystyle 0.0024$ M acetic acid is $\displaystyle 7 \cdot 2 \times 10^{-5} \mathrm{~S} \mathrm{~cm}^{-1}$. If $\displaystyle \mathbf{\Lambda}_{\mathrm{m}}^{\circ}$ for acetic acid is $\displaystyle 390 \cdot 5 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$, then calculate the degree of dissociation ( $\displaystyle \alpha$ ) of acetic acid.
(ii)
Calculate the cell potential for the following half cell reaction at $\displaystyle 25$°C : \[\mathrm{Ag}^{+}(\mathrm{aq})+1 \mathrm{e}^{-} \longrightarrow \mathrm{Ag}(\mathrm{~s}) \] Given that : $\displaystyle \left[\mathrm{Ag}^{+}\right]=0.01 \mathrm{M}$ and \[\mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{\mathrm{o}}=+0.80 \mathrm{~V} \] $\displaystyle [\log 10=1]$
(iii)
Write the name of the electrolyte used in :
(I)
Dry cell
(II)
Fuel cell $\displaystyle \left(\mathrm{H}_{2}-\mathrm{O}_{2}\right)$
Marking-scheme solution
(i)
E(cell)=
°
n
log Kc
E(cell)=
°
log($\displaystyle 1015$)
E(cell)=
°
(15log $\displaystyle 10$) ⟹E(cell)=
°
0.059V
E(cell)=
°
$\displaystyle 0.0295$ × $\displaystyle 15$ = $\displaystyle 0.4425$ V
(ii)
(iii)
$\displaystyle 96500$ C
(i)
Λm =
κ
c
Λm =
$\displaystyle 7.2$ x $\displaystyle 10$−$\displaystyle 5$ x $\displaystyle 1000$
Λm = $\displaystyle 30$ Scm2mol−$\displaystyle 1$
α =
Λm
Λm
°
α =
α = $\displaystyle 0.0768$ or $\displaystyle 0.077$
(b)
(ii)
E(Ag+/(Ag) = E(Ag+/(Ag)
θ
−
log
[Ag]
[Ag+]
E(Ag+/(Ag) = $\displaystyle 0.80$ −0.059log
E(Ag+/(Ag) = $\displaystyle 0.80$ −0.059log $\displaystyle 102$
E(Ag+/(Ag) = $\displaystyle 0.80$ −$\displaystyle 0.059$ x $\displaystyle 2$ ⟹E(Ag+/(Ag) = $\displaystyle 0.80$ −$\displaystyle 0.118$
E(Ag+/(Ag) = 0.682V
(b)
(iii)
(I)
NH4Cl and ZnCl2
(II)
Aq NaOH solution/KOH solution
o $\displaystyle 0$ o -
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CBSE Class 12 Chemistry past-paper question from the 2026 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.