- A$Mn$ is reduced from $Mn^{+4}$ to $Mn^{+3}$
- ✓$NH_3$ gas liberated out
- C$Zn$ is used as anode
- DPaste of $NH_4Cl$ and $ZnCl_2$ is used
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Product $(B)$ of above the reaction is
$Image$
Reason : The reaction between nitrogen and oxygen requires high temperature.
$Ag\left( s \right)\left| {AgBr\left( s \right)\,\left| {B{r^ - }\left( {0.01\,M} \right)} \right|\,\left| {{I^ - }\left( {0.02\,M} \right)} \right|\,AgI\left( s \right)} \right|Ag\left( s \right)$
the correct information is
[Given : $K_{sp}\,\left( {AgBr} \right) = 4 \times {10^{ - 13}}$ ,
$K_{sp}\,\left( {AgI} \right)$ $ = 8 \times {10^{ - 17}},\frac{{2.303\,RT}}{F} = 0.06\,V,\,\log \,2 = 0.3]$
$Cl_2 + 2e^-\to 2CI^{\Theta }$ ; $E^o = 1.36\, V$
$Mn^{+3} + e^-\to Mn^{+2}$ ; $E^o = 1.50\, V$
$Fe^{+3} + e^-\to Fe^{+2}$ ; $E^o = 0.77\, V$
Which of the following is a correct statement
(Assume $100\%$ ionization)
$A.$ $0.500\,M\,C _2 H _5 OH ( aq )$ and $0.25\, M\, KBr ( aq )$
$B.$ $0.100\,M\,K _4\left[ Fe ( CN )_6\right]$ (aq) and $0.100\, M$ $FeSO _4\left( NH _4\right)_2 SO _4$ (aq)
$C.$ $0.05 \,M\, K _4\left[ Fe ( CN )_6\right]( aq )$ and $0.25\, M\, NaCl$ (aq)
$D.$ $0.15\, M\, NaCl ( aq )$ and $0.1\, M BaCl _2$ (aq)
$E.$ $0.02\, M\, KCl\, MgCl _{2 .} 6 H _2 O ( aq )$ and $0.05\, M$ $KCl ( aq )$