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+1 vote
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in Chemistry by (67.5k points)
Given: `Pt(s)"|"underset(1"bar")H_2(g)"|"0.1MNH_4OH(aq)"||"0 .1MCH_3COOH(aq)"|"underset(1"bar")H_2(g)"|"Pt(s)`
`pK_b(NH_4OH)=5, (CH_3COOH)=5,(2.303RT)/F=0.06`
Volume of 0.1 m `NH_4OH` in anode half cell=100ML
Volume of 0.1 M `CH_3COOH` in cathode half cell =100mL
Which is /are correct statement?
A. The emf of given cell is 0.48V.
B. The emf of given cell is 0.36V when 50mL,0.1M NaOH added to cathode compartment
C. The emf of given cell is 0.36V when 50mL 0.1M HCl added to anode compartment
D. The emf of given cell is 0.192V when 100mL 0.1M NaOH added to anode compartment

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1 Answer

0 votes
by (47.2k points)
Correct Answer - A::B::C

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