United States presidential election in Rhode Island, 1896

Main article: United States presidential election, 1896

United States presidential election in Rhode Island, 1896

November 3, 1896

Nominee William McKinley William Jennings Bryan Party Republican Democratic Home state Ohio Nebraska Running mate Garret Hobart Arthur Sewall Electoral vote 4 0 Popular vote 37,437 14,459 Percentage 68.33% 26.39%

President before election Grover Cleveland Democratic Elected President William McKinley Republican

The 1896 United States presidential election in Rhode Island took place on November 3, 1896. Voters chose four representatives, or electors to the Electoral College, who voted for President and Vice President.

Rhode Island voted for the Republican nominee, former governor of Ohio William McKinley, over the Democratic nominee, former U.S. Representative from Nebraska William Jennings Bryan. McKinley won the state by a wide margin of 41.94%.

Results

United States presidential election in Rhode Island, 1896
Party		Candidate	Running mate	Popular vote		Electoral vote
				Count	%	Count	%
	Republican	William McKinley of Ohio	Garret Hobart of New Jersey	37,437	68.33%	4	100.00%
	Democratic	William Jennings Bryan of Nebraska	Arthur Sewall of Maine	14,459	26.39%	0	0.00%
style="background-color: Template:National Democratic Party (United States)/meta/color; width: 5px;" |	[[National Democratic Party (United States)|Template:National Democratic Party (United States)/meta/shortname]]	John McAuley Palmer of Illinois	Simon Bolivar Buckner of Kentucky	1,166	2.13%	0	0.00%
	Prohibition	Joshua Levering of Maryland	Hale Johnson of Illinois	1,160	2.12%	0	0.00%
	Socialist Labor	Charles Horatio Matchett of New York	Matthew Maguire of New Jersey	558	1.02%	0	0.00%
	N/A	Others	Others	5	0.01%	0	0.00%
Total				54,785	100.00%	4	100.00%