Answer :
Answer:
The correct option is: 1.5 × 10⁻⁴¹
Explanation:
Given half reactions:
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O, E° = 1.51 V
Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O, E° = 1.33 V
For a Galvanic cell, the standard cell potential: E°cell > 0
E°cell = E°cathode - E°anode = + 1.51 V - 1.33 V = + 0.18 V
∴ CATHODE: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
ANODE: 2Cr³⁺ + 7H₂O → Cr₂O₇²⁻ + 14H⁺ + 6e⁻
Now, multiply cathode reaction with 6 and multiply anode reaction with 5.
CATHODE: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O] × 6
ANODE: 2Cr³⁺ + 7H₂O → Cr₂O₇²⁻ + 14H⁺ + 6e⁻] × 5
Now adding the two equations, to obtain the overall reaction:
6 MnO₄⁻ + 10Cr³⁺ + 11 H₂O → 6Mn²⁺ + 5Cr₂O₇²⁻ + 22 H⁺
Therefore, the reaction quotient is:
[tex]Q = \frac{[Mn^{2+}]^{6}[Cr_{2}O_{7}^{2-}]^{5}[H^{+}]^{22}}{[Cr^{3+}]^{10}[MnO_{4}^{-}]^{6}}\\[/tex]
Given: [MnO₄⁻] = 0.10 M, [Cr³⁺] = 0.40 M, [Mn²⁺] = 0.20 M, [Cr₂O₇²⁻] = 0.30 M, [H⁺] = 0.010 M
Now putting these given values of concentration of species in the above equation, we get;
[tex]Q = \frac{[0.20 M]^{6}[0.30 M]^{5}[0.010 M]^{22}}{[0.40 M]^{10}[0.10 M]^{6}} = 1.48 \times 10^{-41} \approx 1.5 \times 10^{-41}[/tex]
Therefore, the Reaction quotient Q for the given cell reaction = 1.5 × 10⁻⁴¹