Question

Calculate the Gibbs Free Energy change for reaction below when initial concentrations of Cr 3+ and Cu 2+ are 1.00 M.

**** When do I know to use:

E•cell = E•cathode - E•anode

Vs.

E•cell = E•cathode + E•anode

CARPTER 17 Homtuwot 3+ -0.14 V anode 0.337 V 1.077v n his suahan th more poihve Eocele valves

Answer 1

Ans. #8. Follow this simplified approach that eliminates any chances of confusion

#I. Identify the oxidation and reduction half reactions. Write them separately.

#II. The standard redox table gives “REDUCTION potential”. So, the numeric sign of reduction-half reaction remains the same as in the table.

#III. Change the numeric sign of “OXIDATION” half reaction.

#IV. ADD the two E0 values of oxidation and reduction half reactions without any further change in the numeric sign.

# Solution:

Step 1: The net reaction can be written as the sum of following two half-reactions: 3 Cu 2+6e E -0.337 v E0.744 v Net Reaction: 3 Cu 2 Cr3+ 3 Cu2+ + 2 Cr AE -1.081v Oxidation half: Reduction half: + 2 Cr3 6e 2 Cr Step 2: The standard Gibb's free energy change for the reaction is given by - Equation 1 Where, n number of electrons transferred Faraday's constant - F 96.5 kJ mol1 vl Putting the values in equation 1 6 96.5 kJ mol1 v1 x 1.081 v Hence, AG 625.899 kJ mol

# Correct option- a. 623 kJ (closest value)

Answer 2

If the cathode and anode values have the e^- on the standard (left) side of the half cell equations subtract:

     Cr³⁺_(aq) + 3e^-   ==>   Cr_(s)        (Cathode)  E^o = -0.744 v

     Cu²⁺_(aq) + 2e^-   ==>   Cu_(s)       (Anode)  E^o = +0.34 v

E^o_cell = E^o_cathode - E^o_anode

E^o_cell = -0.744 v - 0.34 v

E^o_cell = -1.084 v

If the cathode and anode values have the e^- on the properly represented side of the equations add:

     Cr³⁺_(aq) + 3e^-   ==>   Cr_(s)        (Cathode)  E^o = -0.744 v

     Cu_(s)   ==>  Cu²⁺_(aq) + 2e^-        (Anode)  E^o = -0.34 v

E^o_cell = E^o_cathode + E^o_anode

E^o_cell = -0.744 v + (-0.34) v

E^o_cell = -1.084 v

Keep in mind the standard form for half cell data is typically presented with the oxidized substance (lost e^-)  on the left and the reduced substance (gained e^-) on the right side of the half cell equation. Because of this, the first equation is generally the one you want to use because by subtracting the two you are already accounting for the sign change that would occur if you used the second method (+0.34 changed to -0.34 when the equation flipped)