Calculate the cell potential, Ecell, of a cell operating with the following reac

Question

Calculate the cell potential, Ecell, of a cell operating with the following reaction at 25?C, in which [MnO4-] = 0.010M, [Br-] = 0.010M, [Mn2+] = 0.15 M, and [H+] = 1.0 M. 2MnO4- (aq) + 10Br- (aq) + 16H+ (aq) ? 2Mn2+ (aq) + 5Br2 (l) + 8H2O (l)

Explanation / Answer

Some shortforms to help you: 1) LEO goes GER (Losing Electrons Oxidation, Gaining Electrons Reduction) 2) OARC (Oxidation = anode, reduction = cathode) Step 1: Write the full equations The two half cells you have given are: 1) Ni/Ni2+ so: Ni --> Ni2+ + 2electrons (e-) 2) Cu/Cu2+ so: Cu --> Cu2+ + 2e- Step 2: Find the Eo values in a table (standard REDUCTION potentials Ni2+ + 2e- --> Ni Eo = -0.257V Cu2+ + 2e- --> Cu Eo = 0.340V Step 3: Calculate the standard cell potential I am assuming here that the teachers wants you to calculate the SPONTANEOUS cell potential (i.e. a positive value) The formula for calculating this is: E cell = E cathode - E anode Using the shortforms I mentioned earlier: Cathode= reduction Anode = oxidation So we have: Anode: Ni --> Ni2+ + 2e- Eo = -0.257V Cathode: Cu2+ + 2e- --> Cu Eo= 0.340V Ecell = Ecath - Eanode = 0.340V - (-0.257V) = 0.597V NOTE: you can flip either of the two equations to make it read as a reduction or oxidation. Just leave the cell potentials as you found them in the table, and use the formula as I did. You want to have a positive value, which means this is the reaction that occurs spontaneously, so look at the Eo values in the table and determine which one MUST be the cathode and which one MUST be the anode, so that you end up with a positive value.

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.