Determine Eo for the following reaction, using the given standard reduction potentials

Question

Determine E^o for the following reaction, using the given standard reduction potentials:

Cr(s) + Mn²⁺(aq) → Cr³⁺(aq) + Mn(s)

E^o for Mn²⁺(aq) = -1.18 V

E^o for Cr³⁺(aq) = -0.74 V

My answer was 0.44 but the correct answer is -0.44

Why is the answer negative???

Answer 1

Cr(s) → Cr³⁺ E = -.78 anode(an)

Mn²⁺ → Mn(s) E = -1.18 cathode(cat)

E_cell = E_cat - E_an = -1.18 - (-.74 )= -.44

Answer 2

Reduction at the cathode: Mn²⁺ + 2e- ===> Mn(s) Eº = -1.18 V

Oxidation at the anode: Cr(s) + ===> Cr³⁺ + 3e- Eº = -0.74 V

Using reduction potentials, the Eº_cell can be found as E_cathode - E_anode = -1.18 - (-0.74)

Eºcell = -1.18 + 0.74

Eºcell = -0.44 V