Year 12 Chemistry – Galvanic Cells
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Calculating EMFs for redox reactions and cells – problem solving strategy.
2Ag+ (aq) + Cu (s)  Cu2+ (aq) + 2Ag (s)
To calculate the EMF of a galvanic cell, we first write the redox half reactions associated with the
cell.
2Ag+ (aq) + 2e-  2Ag (s)
Cu (s)  Cu2+ (aq) + 2eSince the required chemical reaction (above) is made by adding these half reactions, it follows that;
{EMF of complete Reaction} = {EMF of reduction halfie} + {EMF of oxidation halfie}
Therefore, the EMF of the reduction half reaction is simply the electrode potential of the Ag+|Ag
electrode = +0.80V
The oxidation half reaction is the reverse of the standard reduction half reaction for Cu|Cu2+.
Therefore we must reverse the electrode potential for this standard half reaction. Hence, Cu|Cu 2+
electrode = -0.34V.
Now, finding the EMF for the cell:
EMFtotal = EAg + (-ECu)
= (0.80 – 0.34)V
= 0.46V
NOTE: The negative you see in front of the ‘ECu’ is because we reversed this standard half reaction!
Lets try another one!
Calculate the standard EMF of the galvanic cell below:
Ag|Ag+ || Cl-|Cl2, Pt
Which of the species is the stronger oxidant?
Chlorine! Therefore, this is our reduction half reaction.
Cl2 + 2e-  2ClThe E for this is +1.36V
This means that oxidation takes place at the silver electrode.
2Ag  2Ag+ + 2eWe note now that this is the reverse of the standard reduction reaction, so its E is
-0.80V.
Now, finding the EMF for the cell:
EMFtotal = ECl + (-EAg)
= (1.36 – 0.80)V
= 0.56V
Now you try! Attempt questions below. You can also complete Q14.3 in Nelson on page 427.
Questions
1. For each reaction below, calculate the standard EMF, and then decide which direction the
reaction goes – as written or in the reverse direction.
a) Cu2+ + Ni  Ni2+ + Cu
b) 2H+ + Cd  Cd2+ + H2
c) Mg2+ + Co  Co2+ + Mg
d) Fe2+ + Sn  Sn2+ + Fe
e) Cd2+ + Mg  Mg2+ + Cd
f) 2H+ + Hg  Hg2+ + H2
2. Calculate the standard EMF of each of the following cells. State clearly which electrode is
positive.
a) Ni|Ni2+ || Ag+|Ag
b) Cu|Cu2+ || Co2+|Co
c) Sn|Sn2+ || Mg2+|Mg
d) Pt, Cl2|Cl- || Fe2+|Fe
e) Pt|I2, I- || Br-, Br2|Pt
ANSWERS: Q1: a) 0.60V c) -2.09V e) -0.31 V Q2: a) 1.06V (Ag) b) 0.62V (Cu) c) 2.23V (Sn) d) 1.81V (Pt) e) 0.55 V