$\downarrow KOH$
${{K}_{2}}C{{O}_{3}}+{{C}_{6}}{{H}_{5}}N{{H}_{2}}\xleftarrow[2\,KOH]{}{{C}_{6}}{{H}_{5}}NCO+KBr+{{H}_{2}}O$
${{C}_{6}}{{H}_{5}}N{{H}_{2}}+{{(C{{H}_{3}}CO)}_{2}}O\to C{{H}_{3}}CONH{{C}_{6}}{{H}_{5}}+C{{H}_{3}}COOH$
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if $E_{C{u^{ + 2}}/Cu}^o\, = \, + 0.34\,V$
$E_{A{g^ + }/Ag}^o\, = \, + 0.80\,V$
$E_{cell}$ will be
$M{n_{(S)}}|Mn_{(aq)}^{ + 2}(0.4\,M)||Sn_{(aq)}^{ + 2}(0.04\,M)|S{n_{(S)}}$,
Calculate free energy change $(\Delta G)$ at $298\,K$ .
Given :$E_{M{n^{ + 2}}|Mn}^o = \, - \,1.18\,V\,;\,E_{S{n^{ + 2}}|Sn}^o\, = \, - \,0.14\,volt$
$\frac{{2.303\,RT}}{F} = 0.06$
