Question

Identify from following cell reactions that is spontaneous under standard state of conditions.

• A. \(\text{Ca(s)}+\text{Cd}^{2+}\text{(aq)}\rightarrow \text{Ca}^{2+}\text{(aq)}+\text{Cd(s)}\), \([E_{\text{Ca}}^0 = -2.866\) V & \(E_{\text{Cd}}^0 = -0.403\) V\(]\)
• B. \(2\text{Br}^-\text{(s)}+\text{Sn}^{2+}\text{(aq)}\rightarrow \text{Br}_2\text{(l)}+\text{Sn(s)}\), \([E_{\text{Br}}^0 = 1.08\) V & \(E_{\text{Sn}}^0 = -0.136\) V\(]\)
• C. \(2\text{Ag(s)}+\text{Ni}^{2+}\text{(aq)}\rightarrow 2\text{Ag}^+\text{(aq)}+\text{Ni(s)}\), \([E_{\text{Ag}}^0 = 0.799\) V & \(E_{\text{Ni}}^0 = -0.257\) V\(]\)
• D. \(2\text{Au(s)}+\text{Zn}^{2+}\text{(aq)}\rightarrow 2\text{Au}^+\text{(aq)}+\text{Zn(s)}\), \([E_{\text{Au}}^0 = 1.68\) V & \(E_{\text{Zn}}^0 = -0.763\) V\(]\)

Hint:

A cell reaction is spontaneous when its standard cell EMF is positive.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
A cell reaction is spontaneous under standard conditions when \(E^0_{cell} > 0\), because \(\Delta G^0 = -nFE^0_{cell}\) is then negative.

Step 2: Key Formula or Approach:
\[ E^0_{cell} = E^0_{\text{cathode}} - E^0_{\text{anode}} \] The species that is reduced is the cathode, the species that is oxidised is the anode.

Step 3: Check option (A).
Ca is oxidised (anode), \(\text{Cd}^{2+}\) is reduced (cathode). \(E^0_{cell} = -0.403 - (-2.866) = +2.463\) V. Positive, so spontaneous.

Step 4: Check option (B).
\(\text{Br}^-\) is oxidised, \(\text{Sn}^{2+}\) is reduced. \(E^0_{cell} = -0.136 - 1.08 = -1.216\) V. Negative, so not spontaneous.

Step 5: Check option (C).
Ag is oxidised, \(\text{Ni}^{2+}\) is reduced. \(E^0_{cell} = -0.257 - 0.799 = -1.056\) V. Not spontaneous.

Step 6: Check option (D).
Au is oxidised, \(\text{Zn}^{2+}\) is reduced. \(E^0_{cell} = -0.763 - 1.68 = -2.443\) V. Not spontaneous.

Final Answer:
Only the reaction in option (A) has a positive cell potential. \[ \boxed{E^0_{cell} = +2.463\text{ V}} \]