Calculate the cell potential for the reaction as written at 25.00 C, given that [Cr2+] = 0.833 M and [Fe2+] = 0.0140 M.

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Calculate the cell potential for the reaction as written at 25.00 C, given that [Cr2+] = 0.833 M and [Fe2+] = 0.0140 M. Cr(s) + Fe2+(aq) = Cr2+(aq) + Fe(s)

The Correct Answer is

First, write the half-reactions.Anode: Cr(s) = Cr2+ + 2e- Cathode: Fe2+ + 2e- = FeUse the standard reduction potential table.(-0.44 V) – (-0.91 V) = 0.47 VTwo moles of electrons were transferred so n=2.T = 25 + 273.15 = 298.15 KQ = 0.833/0.0140 = 59.5E = E naught – RT/nFln(Q)E = 0.47 – (8.3145)(298.15)/(2)(96485)ln(59.5)E = 0.42 V

Reason Explained

First, write the half-reactions.Anode: Cr(s) = Cr2+ + 2e- Cathode: Fe2+ + 2e- = FeUse the standard reduction potential table.(-0.44 V) – (-0.91 V) = 0.47 VTwo moles of electrons were transferred so n=2.T = 25 + 273.15 = 298.15 KQ = 0.833/0.0140 = 59.5E = E naught – RT/nFln(Q)E = 0.47 – (8.3145)(298.15)/(2)(96485)ln(59.5)E = 0.42 V is correct for Calculate the cell potential for the reaction as written at 25.00 C, given that [Cr2+] = 0.833 M and [Fe2+] = 0.0140 M.