Answer : The value of
is, 0.34 V
Explanation :
Here, copper will undergo reduction reaction will get reduced. Zinc will undergo oxidation reaction and will get oxidized.
The oxidation-reduction half cell reaction will be,
Oxidation half reaction: 
Reduction half reaction: 
Oxidation reaction occurs at anode and reduction reaction occurs at cathode. That means, gold shows reduction and occurs at cathode and chromium shows oxidation and occurs at anode.
The overall balanced equation of the cell is,

To calculate the
of the reaction, we use the equation:


Putting values in above equation, we get:


Hence, the value of
is, 0.34 V
In which elements are belonging to? I will grant that option B ; ti's the best answer.
In which thou might contain the elements zinc , gold , aluminium , and last by not least oxygen.
I truly hope ti's answer helps thou.
The answer is motion, this is what I would go with because when you are dealing with gases it puts motion in the term of particles.
I’m thinking
It conducts electricity when Molten.
<h2>
Answer:</h2>
<h3>Hg2 ( NO2 )2</h3>
<h2>Explanation:</h2>
<h3>Formula of mercury ( 1 ) dioxonitrate 111 is Hg2 ( NO2 )2//</h3>