Identify the cathode and anode, write the net cell equation, and calculate the cell potential for #"Pt"(s) | "SO"_4^(2-)(aq), "H"^(+)(aq) | "H"_2"SO"_3(aq) "||" "Ag"^(+)(aq) | "Ag"(s)#?

There's a small trick here... Ignoring the charge balance, the main action going on in one half-reaction is:

#"SO"_4^(2-)(aq) -> "H"_2"SO"_3(aq) -> "H"_2"O"(l) + "SO"_2(g)#

As a result, we find the two half-reactions to be:

#"Ag"^(+)(aq) + e^(-) -> "Ag"(s)#, #E_(red)^@ = +"0.80 V"#

#"SO"_4^(2-)(aq) + 4"H"^(+)(aq) + 2e^(-) -> "SO"_2(g) + 2"H"_2"O"(l)#, #E_(red)^@ = +"0.20 V"#

The silver reduction has a more positive #E_(red)^@#, so reduction is more spontaneous for #"Ag"^+# and #"Ag"^(+)# is reduced at the cathode.

Thus, we reverse the other half-reaction to write it as an oxidation and the decomposition of sulfurous acid leads to the oxidation of sulfur dioxide to sulfate at the anode.

#E_(o x)^@(2) = -E_(red)^@ (2) = -(+"0.20 V") = -"0.20 V"#

The reactions add up as:

#2("Ag"^(+)(aq) + cancel(e^(-)) -> "Ag"(s))#
#ul(stackrel(color(blue)(+4))("S")"O"_2(g) + 2"H"_2"O"(l) -> stackrel(color(blue)(+6))("S")"O"_4^(2-)(aq) + 4"H"^(+)(aq) + cancel(2e^(-)))#
#color(blue)(2"Ag"^(+)(aq) + "SO"_2(g) + 2"H"_2"O"(l) -> 2"Ag"(s) + "SO"_4^(2-)(aq) + 4"H"^(+)(aq))#

For this reaction, we then get:

#color(blue)(E_(cell)^@) = E_(red)^@ + E_(o x)^@#

#= "0.80 V" + (-"0.20 V") = color(blue)(+"0.60 V")#

And we do not multiply #E_(red)^@# or #E_(o x)^@# by any coefficients since the scaled mols of electrons and scaled mols of reactants/products cancel out.