Answer
See answer below.
Work Step by Step
a) First reaction:
$Cu^+$ is reduced (ox. number: +2 to +1), so it's the oxidizing agent.
$I^-$ is oxidized (ox. number: -1 to -1/3), so it's the reducing agent.
Second reaction:
$I^-$ is reduced (ox. number: -1/3 to -1), so it's the oxidizing agent.
$S_2O_3^{2-}$ is oxidized (S ox. number: +2 to +5/2), so it's the reducing agent.
b) Number of moles of sodium thiosulfate:
$26.32\ mL\times 0.101\ M=2.66\ mmol$
From stoichiometry: $1.33\ mmol\ I_3^-, 2.66\ mmol\ Cu^{2+}$
$2.66\ mmol\times 63.546\ mg/mmol=168.9\ mg$
Mass percentage:
$0.169/0.251\times 100\%=67.3\%$
