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Question

If hydrogen electrode dipped in 2 solution of pH = 3 and pH = 6 and salt bridge is connected the e.m.f. of resulting cell is :


A
0.177 V
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B
0.3 V
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C
0.052 V
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D
0.104 V
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Solution

The correct option is A 0.177 V
Given $$p{ H }_{ 1 }=3$$
$$p{ H }_{ 2 }=6$$
$${ E }_{ cell }=-0.059\log { \cfrac { { 10 }^{ -6 } }{ { 10 }^{ -3 } }  } $$
$$=-0.059\log { { 1 }0^{ -3 } } $$
$$=-0.059\left( -3 \right) =0.177V=$$Emf of the cell

Chemistry
NCERT
Standard XII

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