1896 United States presidential election in Iowa

The 1896 United States presidential election in Iowa took place on November 3, 1896. All contemporary 45 states were part of the 1896 United States presidential election. Iowa voters chose thirteen electors to the Electoral College, which selected the president and vice president.

Main article: 1896 United States presidential election

1896 United States presidential election in Iowa

November 3, 1896

Nominee William McKinley William J. Bryan Party Republican Democratic Home state Ohio Nebraska Running mate Garret Hobart Arthur Sewall Electoral vote 13 0 Popular vote 289,293 223,741 Percentage 55.47% 42.90%

President before election Grover Cleveland Democratic Elected President William McKinley Republican

Iowa was won by the Republican nominees, former Ohio Governor William McKinley and his running mate Garret Hobart of New Jersey.

Results

1896 United States presidential election in Iowa[1]
Party		Candidate	Votes	Percentage	Electoral votes
	Republican	William McKinley	289,293	55.47%	13
	Democratic	William Jennings Bryan	223,741	42.90%	0
	National Democratic	John M. Palmer	4,516	0.87%	0
	Prohibition	Joshua Levering	3,192	0.61%	0
	Socialist Labor	Charles Matchett	453	0.09%	0
	National Prohibition	Charles Bentley	352	0.07%	0
Totals			521,547	100.00%	13
Voter turnout					—

References

1. Dave Leip's U.S. Election Atlas; Presidential General Election Results – Iowa

Notes