Question

When combined with cupric ions (aq, 1M) which will produce the largest standard cell voltage?

a. ) KMnO4(aq)
b. ) MnO2(s)
c. ) MnCl2(aq)
d. ) Mn(s)

im guessing its Mn(s)...but i don't know if im right can someone plz help me? :(

Answer 1

Cu^+2 ions in an electrochemical cell would tend to be reduced to Cu. The half cell potential is +0.159 volts.
MnO4^-, MnO2, and Mn^+2 (in MnCl2) are all oxidizing agents and tend to take electrons from a reducing agent. They could not provide electrons for the reduction of Cu^+2.
The only likely electrochemical reaction is:
Cu^+2(aq) + Mn(s) --> Mn^+2(aq) + Cu(s)
The voltage would be
0.159 + 1.185 = 1.344 volts.
I got those values from the standard reduction potential for Cu^+2 and the standard oxidation potential for Mn (the reverse of the reduction potential for Mn^+2).
No other combination yields a positive voltage larger than 1.344 volts. Does this validate your guess?