A certain metal M forms a soluble nitrate salt MNO3. Suppose the left half cell of a galvanic cell apparatus is filled with a 1.50 M solution of MNO3 and the right half cell with a 15.0 mM 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?

please  see answer below

Explanation:

                         Salt bridge

Anode                                        Cathode

M(s)/MNO3aq)(1.5M)// MNO3(aq)(15nM/ M(s)

MNO3--> M⁺ + NO⁻₃

M(s)->M⁺(aq) +e⁻ Oxidation

The left side anode is placed which is negative due to the less positive  value at the standard electrode potential. Oxidaton occurs here  

M⁺(aq) +e⁻ -> M(s)

The right side cathode is placed which is positive due to the less negative value at the standard electrode potential. Reduction occurs here  

a)For a concentration cell to have positive Ecell,  the concentration of the cathode/reduction half cell must  be greater than anode/oxidation half cell.  Therefore, the right electrode which is the cathode will be positive

b) so i  just added the answer for Ecell, incase you omitted the question.

Ecell =( 2.303×R×T/nF )log([cathode]/[anode])

Cathode  is greater than anode R = 8.314 J/K/mol, T = 273+20 = 293 K

F- Faraday's constant = 96500 C/mol,

n = number of moles of electron= 1

Ecell ={ 2.303×8.314×293 /(1×96500)} log(15/0.015)

note 15mM = 0.015M}

Ecell = 0.116V = 0.12V

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