Q24 / 150 · MHT CET 2026 (13 April Shift 1) · Chemistry
Identify from following cell reactions that is spontaneous under standard state of conditions.
- Correct answerA$\text{Ca(s)} + \text{Cd}^{2+}\text{(aq)} \rightarrow \text{Ca}^{2+}\text{(aq)} + \text{Cd(s)}$, $[E^0_{\text{Ca}} = -2.866$ V & $E^0_{\text{Cd}} = -0.403$ V$]$
- B$2\text{Br}^-\text{(s)} + \text{Sn}^{2+}\text{(aq)} \rightarrow \text{Br}_2\text{(l)} + \text{Sn(s)}$, $[E^0_{\text{Br}} = 1.08$ V & $E^0_{\text{Sn}} = -0.136$ V$]$
- C$2\text{Ag(s)} + \text{Ni}^{2+}\text{(aq)} \rightarrow 2\text{Ag}^+\text{(aq)} + \text{Ni(s)}$, $[E^0_{\text{Ag}} = 0.799$ V & $E^0_{\text{Ni}} = -0.257$ V$]$
- D$2\text{Au(s)} + \text{Zn}^{2+}\text{(aq)} \rightarrow 2\text{Au}^+\text{(aq)} + \text{Zn(s)}$, $[E^0_{\text{Au}} = 1.68$ V & $E^0_{\text{Zn}} = -0.763$ V$]$
Solution
For a cell reaction to be spontaneous under standard conditions, the standard cell potential $E^{\circ}_{\text{cell}}$ must be positive.
$E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}}$
Evaluating the given options:
For $\text{Ca(s)} + \text{Cd}^{2+}\text{(aq)} \rightarrow \text{Ca}^{2+}\text{(aq)} + \text{Cd(s)}$:
Cathode (reduction): $\text{Cd}^{2+} \rightarrow \text{Cd}$
Anode (oxidation): $\text{Ca} \rightarrow \text{Ca}^{2+}$
$E^{\circ}_{\text{cell}} = -0.403 - (-2.866) = +2.463 \text{ V}$ (Spontaneous)
For $2\text{Br}^-\text{(s)} + \text{Sn}^{2+}\text{(aq)} \rightarrow \text{Br}_2\text{(l)} + \text{Sn(s)}$:
Cathode: $\text{Sn}^{2+} \rightarrow \text{Sn}$
Anode: $2\text{Br}^- \rightarrow \text{Br}_2$
$E^{\circ}_{\text{cell}} = -0.136 - 1.08 = -1.216 \text{ V}$ (Non-spontaneous)
For $2\text{Ag(s)} + \text{Ni}^{2+}\text{(aq)} \rightarrow 2\text{Ag}^+\text{(aq)} + \text{Ni(s)}$:
Cathode: $\text{Ni}^{2+} \rightarrow \text{Ni}$
Anode: $\text{Ag} \rightarrow \text{Ag}^+$
$E^{\circ}_{\text{cell}} = -0.257 - 0.799 = -1.056 \text{ V}$ (Non-spontaneous)
For $2\text{Au(s)} + \text{Zn}^{2+}\text{(aq)} \rightarrow 2\text{Au}^+\text{(aq)} + \text{Zn(s)}$:
Cathode: $\text{Zn}^{2+} \rightarrow \text{Zn}$
Anode: $\text{Au} \rightarrow \text{Au}^+$
$E^{\circ}_{\text{cell}} = -0.763 - 1.68 = -2.443 \text{ V}$ (Non-spontaneous)
The only reaction with a positive standard cell potential is the first one.
Answer: $\text{Ca(s)} + \text{Cd}^{2+}\text{(aq)} \rightarrow \text{Ca}^{2+}\text{(aq)} + \text{Cd(s)}$, $[E^0_{\text{Ca}} = -2.866$ V & $E^0_{\text{Cd}} = -0.403$ V$]$
$E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}}$
Evaluating the given options:
For $\text{Ca(s)} + \text{Cd}^{2+}\text{(aq)} \rightarrow \text{Ca}^{2+}\text{(aq)} + \text{Cd(s)}$:
Cathode (reduction): $\text{Cd}^{2+} \rightarrow \text{Cd}$
Anode (oxidation): $\text{Ca} \rightarrow \text{Ca}^{2+}$
$E^{\circ}_{\text{cell}} = -0.403 - (-2.866) = +2.463 \text{ V}$ (Spontaneous)
For $2\text{Br}^-\text{(s)} + \text{Sn}^{2+}\text{(aq)} \rightarrow \text{Br}_2\text{(l)} + \text{Sn(s)}$:
Cathode: $\text{Sn}^{2+} \rightarrow \text{Sn}$
Anode: $2\text{Br}^- \rightarrow \text{Br}_2$
$E^{\circ}_{\text{cell}} = -0.136 - 1.08 = -1.216 \text{ V}$ (Non-spontaneous)
For $2\text{Ag(s)} + \text{Ni}^{2+}\text{(aq)} \rightarrow 2\text{Ag}^+\text{(aq)} + \text{Ni(s)}$:
Cathode: $\text{Ni}^{2+} \rightarrow \text{Ni}$
Anode: $\text{Ag} \rightarrow \text{Ag}^+$
$E^{\circ}_{\text{cell}} = -0.257 - 0.799 = -1.056 \text{ V}$ (Non-spontaneous)
For $2\text{Au(s)} + \text{Zn}^{2+}\text{(aq)} \rightarrow 2\text{Au}^+\text{(aq)} + \text{Zn(s)}$:
Cathode: $\text{Zn}^{2+} \rightarrow \text{Zn}$
Anode: $\text{Au} \rightarrow \text{Au}^+$
$E^{\circ}_{\text{cell}} = -0.763 - 1.68 = -2.443 \text{ V}$ (Non-spontaneous)
The only reaction with a positive standard cell potential is the first one.
Answer: $\text{Ca(s)} + \text{Cd}^{2+}\text{(aq)} \rightarrow \text{Ca}^{2+}\text{(aq)} + \text{Cd(s)}$, $[E^0_{\text{Ca}} = -2.866$ V & $E^0_{\text{Cd}} = -0.403$ V$]$
Hence, the correct answer is option: 'A'.