Question

A student is given the electrochemical cell reaction:MnO4-(aq) + 8H+(aq) + 5e- → Mn2+(aq) + 4H2O(l)Calculate the standard potential (in volts) of the cell at 25°C given that the standard reduction potential of MnO4-(aq)/Mn2+(aq) couple and the standard reduction potential of the H+(aq)/H2(g) couple are +1.51 V and 0.00 V, respectively. The concentration of MnO4-(aq) is 0.010 M and that of H+(aq) is 1.00 M.

Answer 1

To calculate the standard potential of the cell, we can use the Nernst equation:E = E -  RT/nF  * ln Q where E is the cell potential, E is the standard cell potential, R is the gas constant  8.314 J/molK , T is the temperature in Kelvin  298 K , n is the number of electrons transferred in the reaction  5 in this case , F is the Faraday constant  96485 C/mol , and Q is the reaction quotient.First, we need to find the standard cell potential  E . Since the standard reduction potential of MnO4-/Mn2+ couple is +1.51 V and that of H+/H2 couple is 0.00 V, the standard cell potential is:E = E MnO4-/Mn2+  - E H+/H2  = 1.51 V - 0.00 V = 1.51 VNow, we need to find the reaction quotient  Q . The balanced equation is:MnO4- aq  + 8H+ aq  + 5e-  Mn2+ aq  + 4H2O l The reaction quotient Q is given by:Q = [Mn2+]/ [MnO4-][H+]^8 Since we are given the concentrations of MnO4- and H+, we can plug them into the equation:Q = [Mn2+]/ [0.010][1.00]^8 However, we don't have the concentration of Mn2+. To solve this problem, we can assume that the concentration of Mn2+ is very small compared to the concentrations of MnO4- and H+. Therefore, we can approximate Q as:Q  1/ [0.010][1.00]^8  = 1/0.010 = 100Now, we can plug all the values into the Nernst equation:E = 1.51 -  8.314 * 298 /  5 * 96485   * ln 100 E = 1.51 -  0.0171  * ln 100 E = 1.51 -  0.0171  * 4.605E = 1.51 - 0.0787E  1.43 VSo, the standard potential of the cell at 25C is approximately 1.43 volts.