Test 2 Question 2

[Gwen Peng 1L]

Can someone please explain this problem? It asks to find K of Cr^3+(aq)+Fe(s) -> Fe^2+(aq)+Cr(s) and gives the following:
Fe^2+(aq)+2e- -> Fe(aq) E=-0.44 V
Cr^3+(aq)+3e- -> Cr(s) E=-0.74 V

[Anna Goldberg 2I]

First, calculate E standard by E standard=E cathode-E anode
Then, use the equation E standard=(0.05916/n)*log(K) to solve for K.

[Christy Zhao 1H]

Adding on to the comment above, you can also use the equation to solve for K.

[Wenjie Dong 2E]

E_cell=-0.44 V-(-0.74 V)=0.30V, and then use the equation mentioned above to calculate K.

[Gwen Peng 1L]

E_cell=-0.44 V-(-0.74 V)=0.30V, and then use the equation mentioned above to calculate K.

Could you explain why you used -0.44 V as the cathode value if Fe is being oxidized since it gains electrons? Shouldn't -0.74 V be the cathode value since Cr is being reduced since it lost electrons?