Question

0.10 M MnO4– 0.40 M Cr3+ 0.20 M Mn2+ 0.30 M Cr2O72– 0.010 M H+ 0.010 M H+ The standard reduction potentials are as follows: MnO4– + 8H+ + 5e– → Mn2+ + 4H2O, ε° = 1.51 V Cr2O72– + 14H+ + 6e– → 2Cr3+ + 7H2O, ε° = 1.33 V What is the value of Q, the reaction quotient, for this cell reaction?

6.7 × 10^40
1.5 × 10^–41
6.7 × 10^3
1.5 × 10^–4
None of these choices are correct.

Answer 1

The correct option is: 1.5 × 10⁻⁴¹

Step-by-step 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:

`Q = ([Mn^(2+)]^(6)[Cr_(2)O_(7)^(2-)]^(5)[H^(+)]^(22))/([Cr^(3+)]^(10)[MnO_(4)^(-)]^(6))\\`

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;

`Q = ([0.20 M]^(6)[0.30 M]^(5)[0.010 M]^(22))/([0.40 M]^(10)[0.10 M]^(6)) = 1.48 * 10^(-41) \approx 1.5 * 10^(-41)`

Therefore, the Reaction quotient Q for the given cell reaction = 1.5 × 10⁻⁴¹