CBSE 2026 · Region 3 · Set 1 · Q33 · 5 marks
(i)Calculate the electrode potential of a half-cell for zinc electrode dipping in $\displaystyle 0.01$ $\displaystyle \mathrm{M} \mathrm{ZnSO}_{4}$ solution at $\displaystyle 25$°C. Given : $\displaystyle \mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\mathrm{o}}=-0.76 \mathrm{~V}$ $\displaystyle [\log 10=1]$(ii)Write anode, cathode and overall reaction involved in dry cell.(iii)Equilibrium constant $\displaystyle \left(\mathrm{K}_{\mathrm{c}}\right)$ is related to $\displaystyle \mathrm{E}_{\text {cell }}^{0}$, but not to $\displaystyle \mathrm{E}_{\text {cell }}$. Why ? \[\](i)The conductivity of $\displaystyle 0.001$ M solution of acetic acid is $\displaystyle 3.905 \times 10^{-5} \mathrm{~S} \mathrm{~cm}^{-1}$. Calculate its molar conductivity and degree of dissociation ( $\displaystyle \alpha$ ). Given : $\displaystyle \lambda_{\mathrm{CH}_{3} \mathrm{COO}^{-}}^{\mathrm{o}}=40 \cdot 9 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$ \[\lambda_{\mathrm{H}^{+}}^{\mathrm{o}}=349 \cdot 6 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1} \](ii)Give reasons for the following :(I)Why does a mercury cell deliver a constant voltage for its entire life ?(II)Why is it necessary to use salt bridge in a galvanic cell ?
(i)
Calculate the electrode potential of a half-cell for zinc electrode dipping in $\displaystyle 0.01$ $\displaystyle \mathrm{M} \mathrm{ZnSO}_{4}$ solution at $\displaystyle 25$°C. Given : $\displaystyle \mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\mathrm{o}}=-0.76 \mathrm{~V}$ $\displaystyle [\log 10=1]$
(ii)
Write anode, cathode and overall reaction involved in dry cell.
(iii)
Equilibrium constant $\displaystyle \left(\mathrm{K}_{\mathrm{c}}\right)$ is related to $\displaystyle \mathrm{E}_{\text {cell }}^{0}$, but not to $\displaystyle \mathrm{E}_{\text {cell }}$. Why ? \[\]
(i)
The conductivity of $\displaystyle 0.001$ M solution of acetic acid is $\displaystyle 3.905 \times 10^{-5} \mathrm{~S} \mathrm{~cm}^{-1}$. Calculate its molar conductivity and degree of dissociation ( $\displaystyle \alpha$ ). Given : $\displaystyle \lambda_{\mathrm{CH}_{3} \mathrm{COO}^{-}}^{\mathrm{o}}=40 \cdot 9 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$ \[\lambda_{\mathrm{H}^{+}}^{\mathrm{o}}=349 \cdot 6 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1} \]
(ii)
Give reasons for the following :
(I)
Why does a mercury cell deliver a constant voltage for its entire life ?
(II)
Why is it necessary to use salt bridge in a galvanic cell ?
Marking-scheme solution
(i)
zn
E
/Zn
E Zn
/ Zn
log
Zn
=
−
[
]
[
]
[
]
zn
E
/Zn
0.76V
log
=−
−
zn
E
/Zn
= −
−
×
V
(
)
(
)
(
)
= − 0.819V
(ii)
Overall → Zn + 2MnO2 + 2NH4
+ → Zn2+ + 2MnO(OH) + 2NH3
(iii)
Because
cell
E
is a constant value, whereas Ecell becomes zero at
equilibrium.
(i)
m
CH COO
H
−
Λ
=•
+•
= ($\displaystyle 349.6$ + $\displaystyle 40.9$) Scm2 mol−$\displaystyle 1$
= $\displaystyle 390.5$ Scm2 mol−$\displaystyle 1$
m
K $\displaystyle 1000$ Scm mol
C
−
Λ
=
×
−
×
=
×
= $\displaystyle 39.05$ Scm2 mol−$\displaystyle 1$
m
m
Λ
α=
Λ
=
=
(ii)
(I)
Because overall reaction does not contain any ion whose
concentration can change during its life time.
(II)
Because salt bridge completes the circuit / maintains
electrical neutrality.
o $\displaystyle 0$ o -
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CBSE Class 12 Chemistry past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.