Answer:To calculate the cell voltage, we can use the Nernst equation:
E = E° - (RT/nF) ln Q
Where E is the cell voltage, E° is the standard cell voltage, R is the gas constant, T is the temperature in Kelvin, n is the number of moles of electrons transferred in the cell reaction, F is the Faraday constant, and Q is the reaction quotient.
First, we need to convert the temperature to Kelvin by adding 273.15 to 25.0 °C. This gives us a temperature of 298.15 K.
Next, we need to calculate the standard cell voltage, E°. We can do this by looking up the standard reduction potentials for the half-reactions in the cell. The reduction potential for the half-reaction MnO₂ (s) + 4H+ (aq) + 2e- → Mn²+ (aq) + 2H₂O (1) is 1.51 V, and the reduction potential for the half-reaction 2Cr²+ (aq) + 2e- → 2Cr³+ (aq) is -0.41 V. Since the half-reaction with the more positive reduction potential is the oxidation half-reaction and the half-reaction with the more negative reduction potential is the reduction half-reaction, we can use the difference between the two to calculate the standard cell voltage:
E° = 1.51 V - (-0.41 V) = 1.92 V
Next, we need to calculate the reaction quotient, Q. We can do this by using the equilibrium concentrations of the reactants and products in the cell reaction. The concentration of MnO₂ is not given, so we will assume it is present in large excess and therefore does not change during the reaction. The concentration of H+ in the left half-cell is 1.17 M, and the concentration of H+ in the right half-cell is 2.96 M + 3.39 M = 6.35 M. The concentration of Cr²+ in the left half-cell is 2.96 M, and the concentration of Cr³+ in the right half-cell is 1.31 M. Therefore, the reaction quotient is:
Q = (2.96 M Cr³+)(1.17 M H+)(6.35 M H+)/(2.96 M Cr²+)(1.31 M Cr³+)
= (3.47)(6.35 M H+)/(2.96 M)
= 10.34 M
Step-by-step explanation: