Hello Smol,
When determining whether a cell is galvanic or electrolytic, you have to use a standard reduction potential chart. A galvanic cell is a spontaneous cell, doesn't need an external energy source and has a (+)E value. An electrolytic cell is a nonspontaneous cell, needs an external energy source and has a (-)E value. To use the table, you must find the half reactions and add their reduction potential values based on how the reaction is written.
a) Cr3+(aq) + Co(s) --> Cr2+(aq) + Co 2+
Red. Half rxn: Cr3+(aq) + e- --> Cr2+(aq) E= -0.5
Oxid. Half rxn: Co(s) --> Co2+ + 2e- E= 0.28
Notice how, on the actual reduction potential chart, the value for Co(s) was written the other way
(Co2+ + 2e- ---> Co(s)) and the value was -0.28. Because, in the reaction given, it was flipped we must also flip the half reaction to match AND also flip the E value. Remember, whenever you flip a chemical equation, you must also flip any of its associated values (E, G, H, etc.) and their signs. However, when multiplying redox reactions by any coefficients, you DO NOT have to multiply their corresponding E values like how you would with G values for instance.
After doing this, we can now add the values together (-0.5 + 0.28= -0.22). Because we get an overall negative E value, the reaction is nonspontaneous and we can conclude that this is an electrolytic cell.
b) Ca(s) + Cl2(g) --> Ca2+(aq) + Cl-(aq)
Oxid. Half rxn: Ca(s) ----> Ca2+(aq) E= 2.76
Red. Half rxn: Cl2(g) ---> Cl-(aq) E= 1.36
Total E= 4.12, spontaneous, galvanic
c) Cd(s) + Ni2+(aq) ---> Cd2+(aq) + Ni(s)
Oxid. Half rxn: Cd(s) ---> Cd2+(aq) E= 0.4
Red. Half rxn: Ni2+(aq) --> Ni(s) E= -0.23
Total E= 0.17, spontaneous, galvanic
d) H2O(l) --> H+(aq) + OH-(aq)
This reaction is on the reduction table. Its E value is -0.83 meaning that this reaction is nonspontaneous and electrolytic.
I hope this helps!
