A certain metal M forms a soluble nitrate salt M (NO3)3. Suppose the left half cell of a galvanic cell apparatus is filled with a 25.0 mM solution of M (NO3)3 and the right half cell with a 2.50 M solution of the same substance. Electrodes made of M are dipped into both solutions and a voltmeter is connected between them. The temperature of the apparatus is held constant at 20.0 °C. Which electrode will be positive? O left Be sure your answer has a unit symbol, if necessary, and round it to 2 significant digits. Oright What voltage will the voltmeter show? Assume its positive lead is connected to the positive electrode. 0 X

Given:Left-half cell concentration =25.0 mMRight-half cell concentration =2.50 MThe electrode and the electrolytes are the same in both compartments of the cell. They differ only in the concentration of electrolytes. Therefore, it is a concentration cell.A galvanic cell, also known as a voltaic cell, is an electrochemical cell that generates electrical energy through a spontaneous chemical reaction.At the anode (left electrode) Oxidation half reaction (25.0 mM)At cathode (right electrode) Reduction half reaction (2.50 M)At low concentrations a half-cell oxidation reaction takes place, so it acts as an anode and the electrode of this half-cell is negative. At high concentrations half-cell reduction reaction takes place, so it acts as a cathode and the electrode of this cell is positive. The right side cell is high concentration cell, so it acts as a cathodic half-cell and the electrode is a positive The overall reaction in the concentration cell is the dilution of the electrolyte. The overall reaction is given below.To calculate the voltage use the Nernst equation: ...(1)Where,n = number of electrons involved. M is oxidized to M3+ at the anode and M3+ is reduced to M at the cathode.In both processes, 3 electrons are involved. So, n is 3. is the standard cell potentialF = Faraday's constant R = Universal gas constant T= temperatureDetermine the value of Since both the electrodes are the same, we have Now substitute the values in equation and we get:Therefore, the value of Ecell is 0.089 V.The right electrode is positive.The voltmeter shows a voltage of 0.089 V