For the redox reaction?

Question

Ag(s) + SO42 (aq) + H+(aq) ───────> Ag+(aq) + SO2(g) + H2O(l)

(a) write the oxidation and reduction half reactions; (b) balance the overall equation;

(c) identify the species that is oxidised; (d) identify the species that is reduced.

please help

Answer 1

Ag(s) is the "Species Oxidized", therefore it is the "Reducing Agent"

SO42-(aq) is the "Species Reduced", therefore it is the "Oxidizing Agent"

Balanced Equation:
2Ag(s) + SO42-(aq) + 4H+(aq) ----> 2Ag+(aq) + SO2(g) + 2H2O(l)

Oxidation:
2Ag(s) ----> 2Ag+(aq) + 2e-

Reduction:
SO42-(aq) ----> SO2(g) + O2 + 2e-

Net Reduction:
SO42-(aq) ----> SO2(g) + O2 + 2e-
4H+(aq) ----> 2H2O(l) - O2 - 4e-
SO42-(aq) + 4H+(aq) ----> SO2(g) + 2H2O(l) - 2e-

Net Reaction:
2Ag(s) + SO42-(aq) + 4H+(aq) ----> 2Ag+(aq) +2e- + SO2(g) + O2 + 2e- + 2H2O(l) - O2 - 4e-
which cancels out to be...
2Ag(s) + SO42-(aq) + 4H+(aq) ----> 2Ag+(aq) + SO2(g) + 2H2O(l)
which equals the balanced equation (if I balanced it right)

....I think....I could be wrong because I'm assuming that when you typed SO42, you meant SO42-

yeah......

Answer 2

Reducing agent (which is oxidised): Ag .....> Ag+ + e-

Oxidising agent (which is reduced): (SO4)2- + 4H+ + 2e- -----> SO2 + 2H2O

Now double the first equation, and add them together.

Answer 3

(a) S^6+ = S^4+ + 2e-
2Ag + 2e- = 2Ag+

(b) 2Ag + SO4^2- + 4H+ -> 2Ag+ + SO2 + 2H2O

(c) Ag is oxidized

(d) S is reduced

Answer 4

a)
Ag ---> Ag+
(SO4)2 ---> SO2

b)
8Ag --> 8Ag+ + 8e-
8e- + 8H+ +(SO4)2 --->2SO2 + 4H2O
combine and cancel
8Ag + 8H+ + (SO4)2 --> 8Ag+ +2SO2 + 4H2O

c)
Ag

d)
SO4