Given that Ksp for Agl is 8.51 x 10 ", and E°Ag+/Ag = +0.80V, calculate E0 for the reduction step:
Agl(s) + e– ⇌ Ag(s) + I–(aq)
Hence confirm the statement in Section 8.3 that reduction of silver(l) when in the form of solid Agl is thermodynamically less favourable than reduction of AgCl.
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