Using the cell voltage measured for the first cell studied, with cell chemistry Zn/Zn^2+ \ Cu^2+/Cu, and the known half life potential for Zn^2/Zn calculate the reduction potential for Cu^2+/Cu and enter value below. The information received for this problem were the values obtained during an online lab:

Cu xM cell voltage:1.100 V

Range: 0.005 V

Temp: 25 degrees C

The reduction potential of   is  

Explanation:

From the question we are told that

      The potential of the cell is  

       The range is  

        The temperature is  

Note the reason why Zn is oxized and  Cu is reduced is because Zn is higher than Cu on the electrochemical series

    The reaction at the anode is  

The

    The reaction at the anode is  

Now

substituting values

Hence the reduction potential of   is  

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