Electrochemistry Part 1

Section: Chemistry
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Question : 16 of 100
 
Marks: +1, -0
Emf of the following cell at 298K in V is x×10−2,Zn|Zn2+(0.1M)|Ag+(0.01M)|Ag The value of x is .......... .
(Rounded off to the nearest integer).
[Given, EZn2+∕Zn∘=−0.76V,EAg+∕Ag∘=+0.80V, ‌
2.303RT
F
=0.059
]
[26 Feb 2021 Shift 2]
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