Balance the following chemical equations:
Dear Student
Oxidation is loss of electrons (electrons are added on right hand side of equation)
Reduction is gain of electrons (electrons are added on left hand side of the equation)
(i) First reaction is not an redox reaction. Hence cannot be balanced by ion electron method
oxidation states of Mn, H or S is unchanged
(ii) oxidation: Cu(s)
Cu+(aq) + e
reduction: Fe+3(aq) + e Fe+2(aq)
net reaction: Cu(s) + Fe+3(aq) Cu+(aq) + Fe+2(aq)
(iii) oxidation: Cu(s) Cu+2(aq) + 2e
reduction: [Ag+(aq) + e Ag(s) ] x2 We have to multiply by 2 to cancel out electrons in oxidation half reaction
net reaction: Cu(s) + 2Ag+(aq) Cu+(aq) + 2Ag(s)
(iv) oxidation: Al(s) Al+3(aq) + 3e] x2 We have to cross-multiply to cancel out electrons in both half reaction
reduction: 2H+(aq) + 2e H2(g) ] x3
net reaction: 2Al(s) + 6H+(aq) 2Al+3(aq) + 3H2(g)
(v) oxidation: Ca(s) Ca+2(aq) + 2e
reduction: 2H+(aq) + 2e H2(g)
net reaction: Ca(s) + 2H+(aq) Ca+2(aq) + H2(g)
Regards
Oxidation is loss of electrons (electrons are added on right hand side of equation)
Reduction is gain of electrons (electrons are added on left hand side of the equation)
(i) First reaction is not an redox reaction. Hence cannot be balanced by ion electron method
oxidation states of Mn, H or S is unchanged
(ii) oxidation: Cu(s)
reduction: Fe+3(aq) + e
net reaction: Cu(s) + Fe+3(aq)
(iii) oxidation: Cu(s)
reduction: [Ag+(aq) + e
net reaction: Cu(s) + 2Ag+(aq)
(iv) oxidation: Al(s)
reduction: 2H+(aq) + 2e
net reaction: 2Al(s) + 6H+(aq)
(v) oxidation: Ca(s)
reduction: 2H+(aq) + 2e
net reaction: Ca(s) + 2H+(aq)
Regards