Need help with this electrochemical cell problem. Consider the electrochemical cell described below.ð´ð‘”(ð‘ )|ð´ð‘”ð¶ð‘™(ð‘ )|ð¾ð¶ð‘™(ð‘Žð‘ž, ð‘ ð‘Žð‘¡ð‘¢ð‘Ÿð‘Žð‘¡ð‘’ð‘‘)||ð‘ð‘’ð‘™ð‘™ ð‘ ð‘œð‘™ð‘¢ð‘¡ð‘–ð‘œð‘›|ð¶ð‘¢(ð‘ )The cell solution was made by mixing:25.0 mL of 4.00 mM ð¾ð¶ð‘25.0 mL of 4.00 mM ð¾ð¶ð‘¢(ð¶ð‘)225.0 mL of 0.400 M acid, ð»ð´, with pK= 9.5025.0 mL of ð¾ð‘‚ð» solution The measured voltage was -0.440 V. Calculate the molarity of the KOH solution. Assume that essentially all of the copper(I) is in the form ð¶ð‘¢(ð¶ð‘)2 –. For the right side (the cathode), the half cell reaction is:ð¶ð‘¢(ð¶ð‘)2– + e – ⇌ ð¶ð‘¢(ð‘ ) + 2ð¶ð‘–ð¸Â° = −0.429 V