An electrochemical cell is created with the following setup... What is the E of this cell?

Question

An electrochemical cell is created with the following setup:

 

K|K⁺(1.88M)||Cr³⁺(0.054M)|Cr

 

What is the E of this cell?

Answer 1

K(s) ==> K⁺ + e- is the anode (oxidation)

Cr³⁺ + 3e- ==> Cr(s) is the cathode (reduction)

Looking up standard reduction potentials for each, I find the following:

K⁺ + e- ==> K Eº = -2.92 V

Cr³⁺ + 3e- ==> Cr Eº = -0.74 V

Eºcell = Cathode - Anode = -0.74 - (-2.92) = 2.18 V